Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 24 - The Circle

Answer:

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

(i) Here, h = −2, k = 3 and a = 4

∴ Required equation of the circle:
${\left(x+2\right)}^2+{\left(y-3\right)}^2=4^2$
$\Rightarrow {\left(x+2\right)}^2+{\left(y-3\right)}^2=16$

(ii) Here, h = a, k = b and radius = $\sqrt{a^2+b^2}$

∴ Required equation of the circle:
${\left(x-a\right)}^2+{\left(y-b\right)}^2=a^2+b^2$
$\Rightarrow x^2+y^2-2ax-2by=0$

(iii) Here, h = 0, k = −1 and radius = 1

∴ Required equation of the circle:
${\left(x-0\right)}^2+{\left(y+1\right)}^2={\left(1\right)}^2$
$\Rightarrow x^2+y^2+2y=0$

(iv) Here, h = $a\cos \alpha$, k = $a\sin \alpha$ and radius = a

∴ Required equation of the circle:
${\left(x-a\cos \alpha \right)}^2+{\left(y-a\sin \alpha \right)}^2={\left(a\right)}^2$
$$\begin{gathered} \Rightarrow x^2+a^2{\cos }^2\alpha -2ax\cos \alpha +y^2+a^2{\sin }^2\alpha -2ay\sin \alpha =a^2 \\ \Rightarrow x^2+a^2\left({\sin }^2\alpha +{\cos }^2\alpha \right)-2ax\cos \alpha +y^2-2ay\sin \alpha =a^2 \\ \Rightarrow x^2+a^2-2ax\cos \alpha +y^2-2ay\sin \alpha =a^2 \\ \Rightarrow x^2+y^2-2ax\cos \alpha -2ay\sin \alpha =0 \end{gathered}$$

(v) Here, h = a, k = a and radius = $\sqrt{2}a$

∴ Required equation of the circle:
${\left(x-a\right)}^2+{\left(y-a\right)}^2={\left(\sqrt{2}a\right)}^2$
$$\begin{gathered} \Rightarrow x^2+a^2-2ax+y^2+a^2-2ay=2a^2 \\ \Rightarrow x^2+y^2-2ay-2ax=0 \end{gathered}$$

Answer:

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

(i) Given:
(x − 1)² + y² = 4

Here, h = 1, k = 0 and a = 2

Thus, the centre is (1, 0) and the radius is 2.

(ii) Given:
(x + 5)² + (y + 1)² = 9

Here, h = −5, k = −1 and radius = 3

Thus, the centre is (−5, −1) and the radius is 3.

(iii) Given:
$x^2+y^2-4x+6y=5$

The given equation can be rewritten as follows:
${\left(x-2\right)}^2+{\left(y+3\right)}^2-4-9=5$
$\Rightarrow {\left(x-2\right)}^2+{\left(y+3\right)}^2=18$

Thus, the centre is (2, −3).
And, radius = $\sqrt{18}=3\sqrt{2}$

(iv) Given:
$x^2+y^2-x+2y-3=0$

The given equation can be rewritten as follows:
${\left(x-\frac{1}{2}\right)}^2+{\left(y+1\right)}^2-\frac{1}{4}-1-3=0$
$\Rightarrow {\left(x-\frac{1}{2}\right)}^2+{\left(y+1\right)}^2=\frac{17}{4}$

Thus, the centre is $\left(\frac{1}{2},-1\right)$  and and the radius is $\frac{\sqrt{17}}{2}$.

Answer:

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

Given:
h = 1, k = 2

∴ Equation of the circle = ${\left(x-1\right)}^2+{\left(y-2\right)}^2=a^2$     ...(1)

Also, equation (1) passes through (4, 6).

∴ ${\left(4-1\right)}^2+{\left(6-2\right)}^2=a^2$
     $$\begin{gathered} \Rightarrow 9+16=a^2 \\ \Rightarrow a=5 \left(\because a>0\right) \end{gathered}$$

Substituting the value of a in equation (1):

${\left(x-1\right)}^2+{\left(y-2\right)}^2=25$

$$\begin{gathered} \Rightarrow x^2+1-2x+y^2+4-4y=25 \\ \Rightarrow x^2-2x+y^2-4y=20 \\ \Rightarrow x^2+y^2-2x-4y-20=0 \end{gathered}$$

Thus, the required equation of the circle is $x^2+y^2-2x-4y-20=0$.

Answer:

The point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 is (0, 0).

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

The point of intersection of the lines x + y + 1 = 0 and x − 2y + 4 = 0 is (−2, 1).

∴ h = −2, k = 1

∴ Equation of the required circle = ${\left(x+2\right)}^2+{\left(y-1\right)}^2=a^2$     ...(1)

Also, equation (1) passes through (0, 0).

∴ ${\left(0+2\right)}^2+{\left(0-1\right)}^2=a^2$
$$\begin{gathered} \Rightarrow 4+1=a^2 \\ \Rightarrow a=\sqrt{5} \left(\because a>0\right) \end{gathered}$$

Substituting the value of a in equation (1):
${\left(x+2\right)}^2+{\left(y-1\right)}^2=5$
$$\begin{gathered} \Rightarrow x^2+4+4x+y^2+1-2y=5 \\ \Rightarrow x^2+4x+y^2-2y=0 \end{gathered}$$

Hence, the required equation of the circle is $x^2+y^2+4x-2y=0$.

Answer:

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

The centre of the required circle lies on the positive direction of the y-axis at a distance 6 from the origin.
Thus, the coordinates of the centre are (0, 6).

∴ h = 0, k = 6

∴ Equation of the circle = ${\left(x-0\right)}^2+{\left(y-6\right)}^2=a^2$     ...(1)

Also, a = 4

Substituting the value of a in equation (1):
${\left(x-0\right)}^2+{\left(y-6\right)}^2=16$

$$\begin{gathered} \Rightarrow x^2+y^2+36-12y=16 \\ \Rightarrow x^2+y^2-12y+20=0 \end{gathered}$$

Hence, the required equation of the circle is $x^2+y^2-12y+20=0$.

Answer:

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

The intersection point of 2x + y = 6 and 3x + 2y = 4 is (8, −10).

The diameters of a circle intersect at the centre.
Thus, the coordinates of the centre are (8, −10).
∴ h = 8, k = −10

Thus, the equation of the required circle is ${\left(x-8\right)}^2+{\left(y+10\right)}^2=a^2$     ...(1)

Also, a = 10

Substituting the value of a in equation (1):
${\left(x-8\right)}^2+{\left(y+10\right)}^2=100$
$$\begin{gathered} \Rightarrow x^2+y^2-16x+64+100+20y=100 \\ \Rightarrow x^2+y^2-16x+20y+64=0 \end{gathered}$$

Hence, the required equation of the circle is $x^2+y^2-16x+20y+64=0$.

Answer:

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

(i) Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.
It is given that the circle passes through the points (6, 0) and (0, 6).

∴ ${\left(6-h\right)}^2+{\left(0-k\right)}^2=6^2$
And, ${\left(0-h\right)}^2+{\left(6-k\right)}^2=6^2$

$$\begin{gathered} \Rightarrow {\left(6-h\right)}^2+{\left(-k\right)}^2=36 \\ \Rightarrow 36+h^2-12h+k^2=36 \\ \Rightarrow h^2+k^2=12h ...\left(1\right) \end{gathered}$$

Also, $h^2+36+k^2-12k=36$
$\Rightarrow h^2+k^2=12k$        ...(2)

From (1) and (2), we get:

$12k=12h\Rightarrow h=k$

∴ From equation (2), we have:
$$\begin{gathered} k^2+k^2=12k \\ \Rightarrow k^2-6k=0 \\ \Rightarrow k\left(k-6\right)=0 \\ \Rightarrow k=6 \left(\because k>0\right) \end{gathered}$$

Consequently, we get:
h = 6

Hence, the required equation of the circle is ${\left(x-6\right)}^2+{\left(y-6\right)}^2=36$ or $x^2+y^2-12x-12y+36=0$.

(ii) Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$ .
It is given that the circle with radius 6 units touches the x-axis at a distance of 5 units from the origin.
∴ a = 6, h = 5

Hence, the required equation is ${\left(x-5\right)}^2+{\left(y-0\right)}^2=6^2$ or $x^2+y^2-10x-11=0$.


(iii) Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.
It is given that the circle touches both the axes.

Thus, the required equation will be $x^2+y^2-2ax-2ay+a^2=0$.

Also, the circle passes through the point (2, 1).

∴$4+1-4a-2a+a^2=0$
$$\begin{gathered} \Rightarrow a^2-6a+5=0 \\ \Rightarrow a^2-5a-a+5=0 \\ \Rightarrow a=1,5 \end{gathered}$$

Hence, the required equation is $x^2+y^2-2x-2y+1=0$ or $x^2+y^2-10x-10y+25=0$.

(iv) Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.
Given:
k = −15, a = 17

The circle passes through the point (0, 0).
∴ Equation of the circle:
  ${\left(0-h\right)}^2+{\left(0-15\right)}^2={\left(17\right)}^2$
⇒ $h=\pm 8$

Hence, the required equation of the circle is ${\left(x-8\right)}^2+{\left(y+15\right)}^2={17}^2$ or ${\left(x+8\right)}^2+{\left(y+15\right)}^2={17}^2$, i.e. $x^2+y^2\pm 16x+30y=0$.

Answer:

It is given that the centre is at the point (3, 4).
Let the equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.
∴ Equation of the required circle = ${\left(x-3\right)}^2+{\left(y-4\right)}^2=a^2$    ...(1)

Also, the circle touches the straight line 5x + 12y − 1 = 0.

$$\begin{gathered} \therefore a=\left|\frac{5\left(3\right)+12\left(4\right)-1}{\sqrt{5^2+{12}^2}}\right|=\left|\frac{62}{13}\right| \\ \Rightarrow a^2=\left|\frac{5\left(3\right)+12\left(4\right)-1}{13}\right|=\frac{3844}{169} \end{gathered}$$

So, from equation (1), we have:
${\left(x-3\right)}^2+{\left(y-4\right)}^2=\frac{3844}{169}$
$\Rightarrow x^2+y^2-6x-8y=\frac{3844}{169}-25$
$\Rightarrow 169\left(x^2+y^2-6x-8y\right)+381=0$

Hence, the required equation of the circle is $169\left(x^2+y^2-6x-8y\right)+381=0$.

Answer:

If the circle lies in the third quadrant, then its centre will be (−a, −a).

The centre lies on x − 2y = 3.
∴ $-a+2a=3\Rightarrow a=3$

∴ Required equation of the circle = ${\left(x+3\right)}^2+{\left(y+3\right)}^2=9$   
                                                   =$x^2+y^2+6x+6y+9=0$             

If the circle lies in the fourth quadrant, then its centre will be (a, −a),
∴ $a+2a=3\Rightarrow a=1$

∴ Required equation of the circle = ${\left(x-1\right)}^2+{\left(y+1\right)}^2=1$
                                                   = $x^2+y^2-2x+2y+1=0$ 

Answer:

Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.
The point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0  is $\left(\frac{-1}{17}, \frac{22}{17}\right)$.
∴ Centre = $\left(\frac{-1}{17},\frac{22}{17}\right)$

Also, the circle passes through the origin.

∴ $a^2={\left(\frac{1}{17}\right)}^2+{\left(\frac{22}{17}\right)}^2=\frac{485}{289}$

Hence, the required equation of the circle is ${\left(x+\frac{1}{17}\right)}^2+{\left(y-\frac{22}{17}\right)}^2=\frac{485}{289}$.                                                  

Answer:

It is given that a circle of radius 4 units touches the coordinate axes in the first quadrant.

Centre of the given circle = (4, 4)

The equation of the given circle is ${\left(x-4\right)}^2+{\left(y-4\right)}^2=16$.

The images of this circle with respect to the line mirrors x = 0 and y = 0. They have their centres at $\left(-4, 4\right) \mathrm{and} \left(4, -4\right),$ respectively.

∴ Required equations of the images =  ${\left(x+4\right)}^2+{\left(y-4\right)}^2=16$ and ${\left(x-4\right)}^2+{\left(y+4\right)}^2=16$

                                                         =  $x^2+y^2+8x-8y+16=0$ and $x^2+y^2-8x+8y+16=0$

Answer:

Case I: The centre lies in first quadrant.


Let the required equation be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

Here, AB = 8 units and L (0, 3)

In $△$CAM:
$\Rightarrow C{A}^2=C{M}^2+A{M}^2$
$$\begin{gathered} \Rightarrow C{A}^2=3^2+4^2 \\ \Rightarrow CA=5 \\ \Rightarrow CL=CA=5 \end{gathered}$$

∴ Coordinates of the centre = $\left(5, 3\right)$
And, radius of the circle = 5
${\left(x-5\right)}^2+{\left(y-3\right)}^2=25$, i.e. $x^2+y^2-10x-6y=-9$


Case II: The centre lies in the second quadrant.


Coordinates of the centre = $\left(-5, 3\right)$
And, radius of the circle= 5
${\left(x+5\right)}^2+{\left(y-3\right)}^2=25$, i.e. $x^2+y^2+10x-6y=-9$

Hence, the equation of the required circle is ${\left(x\pm 5\right)}^2+{\left(y-3\right)}^2=25$, i.e. $x^2+y^2\pm 10x-6y=-9$.

Answer:

Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

The circle passes through the points (0, 3) and (0, −3).

∴ ${\left(0-h\right)}^2+{\left(3-k\right)}^2=a^2$       ...(1)
And, ${\left(0-h\right)}^2+{\left(-3-k\right)}^2=a^2$ ...(2)

Solving (1) and (2), we get:
k=0

Given:
Radius = 5
∴ a² = 25

So, from equation (2), we have:
$h^2+9=25\Rightarrow h=\pm 4$

Hence, the required equation is ${\left(x\pm 4\right)}^2+y^2=25$, which can be rewritten as $x^2\pm 8x+y^2-9=0$.

Answer:

We have Area of circle = 154
$$\begin{gathered} \mathrm{\pi }{r}^2=154 \\ \Rightarrow r^2=49 \end{gathered}$$
The intersection of two lines will give us the centre of the circle.
Solving 2x − 3y = 5 and 3x − 4y = 7 we get
x = 1 and y = −1
Now, the equation of the circle is given by
$$\begin{gathered} {\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y+1\right)}^2=49 \\ \Rightarrow x^2+y^2-2x+2y-47=0 \end{gathered}$$

Answer:

The centre and the radius of the circle x² + y² = 16 are (0, 0) and 4
Now, the perpendicular distance from the centre of the circle to the tangent y = $\sqrt{3}$x + k is equal to the radius of the circle
$$\begin{gathered} \therefore 4=\left|\frac{\sqrt{3}\left(0\right)-0+k}{\sqrt{{\left(\sqrt{3}\right)}^2+1^2}}\right| \\ \Rightarrow \pm 4=\frac{k}{2} \\ \Rightarrow k=\pm 8 \end{gathered}$$

Answer:

Solving 3x + y = 14 and 2x + 5y = 18 we get
x = 4 and y = 2
The radius is equal to the distance between (1, −2) and (4, 2)
$$\begin{gathered} r=\sqrt{{\left(4-1\right)}^2+{\left(2+2\right)}^2} \\ =\sqrt{9+16} \\ =5 \end{gathered}$$
Now, the equation of the circle is given by
$$\begin{gathered} {\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y+2\right)}^2=25 \\ \Rightarrow x^2+y^2-2x+4y-20=0 \end{gathered}$$

Answer:

We have 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0
$\Rightarrow y=\frac{3}{4}x+1 \mathrm{and} y=\frac{3}{4}x-\frac{7}{8}$
Since, the slope of both the lines are equal.
Hence, the both the lines are parallel.
The distance between the parralel lines is given by
$$\begin{gathered} \left|\frac{C_1-C_2}{\sqrt{A^2+B^2}}\right| \\ =\left|\frac{4+{\displaystyle \frac{7}{2}}}{\sqrt{3^2+4^2}}\right| \\ =\left|\frac{{\displaystyle \frac{15}{2}}}{5}\right| \\ =\frac{3}{2} \end{gathered}$$​
Now, the radius is equal to the half of the distance between the parallel lines(diameter of the circle).
Hence, the radius is given by
$\frac{3}{4}$

Answer:

Squaring and adding  $x=\frac{2at}{1+t^2}$ and $y=a\left(\frac{1-t^2}{1+t^2}\right)$, we get
$$\begin{gathered} x^2+y^2={\left(\frac{2at}{1+t^2}\right)}^2+a^2{\left(\frac{1-t^2}{1+t^2}\right)}^2 \\ \Rightarrow x^2+y^2=\frac{4a^2{t}^2+a^2-2a^2{t}^2+a^2{t}^4}{{\left(1+t^2\right)}^2} \\ \Rightarrow x^2+y^2=\frac{a^2+2a^2{t}^2+a^2{t}^4}{{\left(1+t^2\right)}^2} \\ \Rightarrow x^2+y^2=a^2\frac{{\left(1+t^2\right)}^2}{{\left(1+t^2\right)}^2} \\ \Rightarrow x^2+y^2=a^2 \end{gathered}$$
Since, the above equation represents the equation of a circle, hence points (x, y) lies on the circle.

Answer:

Centre of the given circle = $\left(1, 1\right)$
Radius of the given circle = 1

This circle is rolled along the positive direction of the x-axis. When it makes one complete roll, its centre moves horizontally through a distance equal to its circumference, i.e 2$\mathrm{\pi }$.

Thus, the coordinates of the centre of the new circle will be $\left(1+2\mathrm{\pi }, 1\right)$.

Hence, the required equation of the circle is ${\left(\mathrm{x}-1-2\mathrm{\pi }\right)}^2+{\left(y-1\right)}^2=1$.

Answer:

Clearly, the centre of the circle lies on the line 4y = x + 7.

The circle passes through A (−3, 4) and B (5, 4).

The slope of the segment joining A and B is zero.

Therefore, the slope of the perpendicular bisector of AB is not defined.

Hence, the perpendicular bisector of AB will be parallel to the y-axis and will pass through $\left(\frac{-3+5}{2}, \frac{4+4}{2}\right) = \left(1, 4\right)$.

The equation of the perpendicular bisector is $x=1$.

The intersection point of the perpendicular bisector and 4y = x + 7 is $\left(1, 2\right)$.

∴ Centre = $\left(1, 2\right)$

Radius = $\sqrt{{\left(5-1\right)}^2+{\left(4-2\right)}^2}=\sqrt{20}$

Hence, the required equation of the circle is $x^2+y^2-2x-4y-15=0$.

Answer:

According to question, the centre of the required circle lies on the line x + y − 9 = 0.

Let the coordinates of the centre be $\left(t, 9-t\right)$.

Let the radius of the circle be a.

Here, a is the distance of the centre from the line 2x − y + 1 = 0.

$$\begin{gathered} \therefore a=\left|\frac{2t-9+t+1}{\sqrt{2^2+{\left(-1\right)}^2}}\right|=\left|\frac{3t-8}{\sqrt{5}}\right| \\ \Rightarrow a^2={\left(\frac{3t-8}{\sqrt{5}}\right)}^2 ...\left(1\right) \end{gathered}$$

Therefore, the equation of the circle is ${\left(x-t\right)}^2+{\left(y-\left(9-t\right)\right)}^2=a^2$.        ...(2)

The circle passes through (2, 5).

∴ ${\left(2-t\right)}^2+{\left(5-\left(9-t\right)\right)}^2=a^2$
$$\begin{gathered} \Rightarrow {\left(2-t\right)}^2+{\left(5-\left(9-t\right)\right)}^2={\left(\frac{3t-8}{\sqrt{5}}\right)}^2 \left(\mathrm{Using} \left(1\right)\right) \\ \Rightarrow 5\left(2t^2-12t+20\right)=9t^2+64-48t \\ \Rightarrow {\left(t-6\right)}^2=0 \\ \Rightarrow t=6 \end{gathered}$$

Substituting t = 6 in (1):

$a^2={\left(\frac{10}{\sqrt{5}}\right)}^2$

Substituting the values of $a^2$ and t in equation (2), we find the required equation of circle to be ${\left(x-6\right)}^2+{\left(y-3\right)}^2=20$.

Answer:

(i) The given equation can be rewritten as $x^2+y^2+2\left(3\right)x-2\left(4\right)y-24=0$.

∴ Centre = $\left(-3,-4\right)$
And, radius = $\sqrt{{\left(3\right)}^2+{\left(4\right)}^2+24}=\sqrt{49}=7$


(ii) The given equation can be rewritten as $x^2+y^2-\frac{3x}{2}+\frac{5y}{2}-\frac{7}{2}=0$.

∴ Centre = $\left(\frac{3}{4},\frac{-5}{4}\right)$
And, radius = $\sqrt{{\left(\frac{3}{4}\right)}^2+{\left(\frac{-5}{4}\right)}^2+\frac{7}{2}}=\sqrt{\frac{34+56}{16}}=\sqrt{\frac{90}{16}}=\frac{3\sqrt{10}}{4}$


(iii) The given equation can be rewritten as $x^2+y^2+2x\cos \theta +2y\sin \theta -8=0$.

∴ Centre = $\left(-\cos \theta ,-\sin \theta \right)$
And, radius = $\sqrt{{\left(-\cos \theta \right)}^2+{\left(-\sin \theta \right)}^2+8}=\sqrt{1+8}=3$


(iv) The given equation can be rewritten as $x^2+y^2-\frac{2ax}{2}-\frac{2by}{2}=0$.

∴ Centre = $\left(\frac{a}{2},\frac{b}{2}\right)$
And, radius = $\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2}=\frac{1}{2}\sqrt{a^2+b^2}$

Answer:

(i) Let the required circle be $x^2+y^2+2gx+2fy+c=0$.      ...(1)
It passes through (5, 7), (8, 1) and (1, 3).

Substituting the coordinates of these points in equation (1):

$74+10g+14f+c=0$    ...(2)
$65+16g+2f+c=0$      ...(3)
$10+2g+6f+c=0$        ...(4)

Simplifying (2), (3) and (4):

$g=\frac{-29}{6}, f=\frac{-19}{6}, c=\frac{56}{3}$

Equation of the required circle:
    $x^2+y^2-\frac{29x}{3}-\frac{19y}{3}+\frac{56}{3}=0$
$\Rightarrow$$3\left(x^2+y^2\right)-29x-19y+56=0$


(ii) Let the required circle be $x^2+y^2+2gx+2fy+c=0$.      ...(1)

It passes through (1, 2), (3, −4) and (5, −6).

Substituting the coordinates of these points in equation (1):

$5+2g+4f+c=0$              ...(2)
$25+6g-8f+c=0$            ...(3)
$61+10g-12f+c=0$       ...(4)

Simplifying (2), (3) and (4):
$g=-11, f=-2, c=25$

The equation of the required circle is $x^2+y^2-22x-4y+25=0$.


(iii) Let the required circle be $x^2+y^2+2gx+2fy+c=0$.      ...(1)

It passes through (5, −8), (−2, 9) and (2, 1).

Substituting the coordinates of these points in equation (1):

$89+10g-16f+c=0$    ...(2)
$85-4g+18f+c=0$      ...(3)
$5+4g+2f+c=0$          ...(4)
 
Simplifying (2), (3) and (4):
$g=58, f=24, c=-285$

The equation of the required circle is $x^2+y^2+116x+48y-285=0$.

(iv) Let the required circle be $x^2+y^2+2gx+2fy+c=0$.      ...(1)
It passes through (0, 0), (−2, 1) and (−3, 2).

Substituting the coordinates of these points in equation (1):

$c=0$                              ...(2)
$5-4g+2f+c=0$        ...(3)
$13-6g+4f+c=0$     ...(4)

Simplifying (2), (3) and (4):

$g=\frac{-3}{2}, f=\frac{-11}{2}, c=0$

The equation of the required circle is $x^2+y^2-3x-11y=0$.

Answer:

Let the required equation of the circle be $x^2+y^2+2gx+2fy+c=0$.      ...(1)
It is given that the circle passes through (3, −2), (−2, 0).
∴ $13+6g-4f+c=0$   ...(2)
$4-4g+c=0$                 ...(3)

The centre lies on the line 2x − y = 3.
∴ $-2g+f-3=0$       ...(4)

Solving (2), (3) and (4):

$g=\frac{3}{2}, f=6, c=2$

Hence, the required equation of circle is $x^2+y^2+3x+12y+2=0$.

Answer:

Let the required equation of the circle be $x^2+y^2+2gx+2fy+c=0$.      ...(1)
It is given that the circle passes through (3, 7), (5, 5).
∴ $58+6g+14f+c=0$   ...(2)
$50+10g+10f+c=0$     ...(3)

The centre lies on the line x − 4y = 1.
∴ $-g+4f-1=0$       ...(4)

Solving (2), (3) and (4):
$g=3, f=1, c=-90$

Hence, the required equation of the circle is $x^2+y^2+6x+2y-90=0$.

Answer:

Let the required equation of the circle be $x^2+y^2+2gx+2fy+c=0$.      ...(1)

It is given that the circle passes through (3, −2), (1, 0), (−1, −2).
∴ $13+6g-4f+c=0$   ...(2)
$1+2g+c=0$                ...(3)
$5-2g-4f+c=0$         ...(4)

Solving (2), (3) and (4):
$g=-1, f=2, c=1$

Theerefore, the equation of the circle is $x^2+y^2-2x+4y+1=0$.      ...(5)

We see that the point (1, −4) satisfies the equation (5).

Hence, the points (3, −2), (1, 0), (−1, −2) and (1, −4) are concyclic.

Answer:

Let the required equation of the circle be $x^2+y^2+2gx+2fy+c=0$.      ...(1)

It is given that the circle passes through (5, 5), (6, 4), (−2, 4).
∴ $50+10g+10f+c=0$   ...(2)
$52+12g+8f+c=0$     ...(3)
$20-4g+8f+c=0$       ...(4)

Solving (2), (3) and (4):
$g=-2, f=-1, c=20$

Thus, the equation of the circle is $x^2+y^2-4x-2y-20=0$.      ...(5)

We see that the point (7, 1) satisfies equation (5).

Hence, the points (5, 5), (6, 4), (−2, 4) and (7, 1) lie on the circle.

Also, centre of the required circle = $\left(2, 1\right)$
Radius of the required circle = $\sqrt{4+1+20}=5$

Answer:

In $∆$ABC:
(i) Let AB represent the line x + y + 3 = 0.       ...(1)
Let BC represent the line x − y + 1 = 0.           ...(2)
Let CA represent the line x = 3.                       ...(3)

Intersection point of (1) and (3) is $\left(3,-6\right)$.
Intersection point of (1) and (2) is (−2, −1).
Intersection point of (2) and (3) is (3, 4).

Therefore, the coordinates of A, B and C are $\left(3,-6\right)$, (−2, −1) and (3, 4), respectively.

Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $45+6g-12f+c=0$
    $5-4g-2f+c=0$
    $25+6g+8f+c=0$

$\therefore g=-3, f=1, c=-15$

Hence, the required equation of the circumcircle is $x^2+y^2-6x+2y-15=0$.


(ii) In $∆$ABC:
Let AB represent the line 2x + y − 3 = 0.    ...(1)
Let BC represent the line x + y − 1 = 0.      ...(2)
Let CA represent the line 3x + 2y − 5 = 0.  ...(3)

Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, −1).
Intersection point of (2) and (3) is (3, −2).

The coordinates of A, B and C are (1, 1), (2, −1) and (3, −2), respectively.

Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $2+2g+2f+c=0$
    $5+4g-2f+c=0$
    $13+6g-4f+c=0$

$\therefore g=\frac{-13}{2}, f=\frac{-5}{2}, c=16$

Hence, the required equation of the circumcircle is $x^2+y^2-13x-5y+16=0$.

(iii) In $∆$ABC:
Let AB represent the line x + y = 2.          ...(1)
Let BC represent the line 3x − 4y = 6.      ...(2)
Let CA represent the line x − y = 0.          ...(3)

Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, 0).
Intersection point of (2) and (3) is (−6, −6).

The coordinates of A, B and C are (1, 1), (2, 0) and (−6, −6), respectively.

Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.
It passes through A, B and C.
∴ $2+2g+2f+c=0$
   $4+4g+c=0$
   $72-12g-12f+c=0$

$\therefore g=2, f=3, c=-12$

Hence, the required equation of the circumcircle is $x^2+y^2+4x+6y-12=0$.

(iv)
In $∆$ABC:
(i) Let AB represent the line y = x + 2          ...(1)
Let BC represent the line 3y = 4x                 ...(2)
Let CA represent the line 2y = 3x                 ...(3)

Intersection point of (1) and (3) is (4, 6)
Intersection point of (1) and (2) is (6, 8).
Intersection point of (2) and (3) is (0, 0).

Therefore, the coordinates of A, B and C are (4, 6), (6, 8) and (0, 0) respectively.

Let the equation of the circumcircle be $x^2+y^2+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $52+8g+12f+c=0$, $100+12g+16f+c=0$
and
 
$$\begin{gathered} 0++0+0+0+c=0 \\ \Rightarrow c=0 \end{gathered}$$

$\therefore g=-23, f=11, c=0$

Hence, the required equation of the circumcircle is $x^2+y^2-46x+22y=0$.

Answer:

The given equations of the circles are as follows:
x² + y² − 4x − 6y − 12 = 0,   ...(1)
x² + y² + 2x + 4y − 10 = 0    ...(2)
And, x² + y² − 10x − 16y − 1 = 0    ...(3)

The centre of circle (1) is (2, 3).
The centre of circle (2) is (−1, −2).
The centre of circle (3) is (5, 8).

The area of the triangle formed by the points (2, 3), (−1, −2) and (5, 8) is $\frac{1}{2}\left|2\left(-10\right)-1\left(5\right)+5\left(5\right)\right| = \frac{1}{2}\left|-25+25\right| = 0$.
Hence, the centres of the circles x² + y² − 4x − 6y − 12 = 0, x² + y² + 2x + 4y − 10 = 0 and x² + y² − 10x − 16y − 1 = 0 are collinear.

Answer:

Let the radii of the circles x² + y² = 1, x² + y² − 2x − 6y − 6 = 0 and x² + y² − 4x − 12y − 9 = 0 be $r_1, r_2 \mathrm{and} r_3$, respectively.

∴ $r_1=1, r_2=\sqrt{{\left(-1\right)}^2+{\left(-3\right)}^2+6}=4, r_3=\sqrt{{\left(-2\right)}^2+{\left(-6\right)}^2+9}=7$

Now, $r_2-r_1=r_3-r_2=3$

∴ $r_1, r_2 \mathrm{and} r_3$ are in A.P.

Answer:

According to the question, the circle passes through the origin.
Let the equation of the circle be $x^2+y^2-2hx-2ky=0$.

The circle cuts off chords of lengths 4 and 6 on the positive sides of the x-axis and the y-axis, respectively.

∴ Centre = $\left(\frac{4}{2}, \frac{6}{2}\right)=\left(2, 3\right)=\left(h, k\right)$
$\Rightarrow h=2, k=3$

∴ Required equation:
$x^2+y^2+2\left(-2\right)x+2\left(-3\right)y=0$
 $\Rightarrow x^2+y^2-4x-6y=0$

Answer:

Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$.
The centre of the circle x² + y² − 6x + 12y + 15 = 0 is (3, −6).

Area of the required circle = $2\mathrm{\pi }{r}^{\mathit{2}}$
Here, r = radius of the given circle

Now, r = $\sqrt{9+36-15}=\sqrt{30}$

∴ Area of the required circle = $2\mathrm{\pi }\left(30\right)=60\mathrm{\pi }$

Let R be the radius of the required circle.

∴ $60\mathrm{\pi }=\mathrm{\pi }{R}^{\mathit{2}}\mathit{\Rightarrow }{R}^{\mathit{2}}\mathit{=}60$

Thus, the equation of the required circle is ${\left(x-3\right)}^2+{\left(y+6\right)}^2=60$, i.e. $x^2+y^2-6x+12y=15$.

Answer:

Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$.

It passes through (1, 1) and (2, 2).
∴ $2g+2f+c=-2$  ...(1)
And, $4g+4f+c=-8$  ...(2)

From (1) and (2), we have:
$-2g-2f=6\Rightarrow g+f=-3$    ...(3)

∴ From (2) and (3), we have:
$c=4$

$$\begin{gathered} \mathrm{Also}, \sqrt{g^2+f^2-c}=1 \\ \Rightarrow g^2+f^2=1+c=5 \\ \Rightarrow {\left(g+f\right)}^2-2gf=5 \\ \Rightarrow gf=2 \end{gathered}$$

Using (3), we get:
$g=-2,-1$

Correspondingly, we have:
$f=-1,-2$

Therefore, the required equations of the circles are $x^2+y^2-4x-2y+4=0$ and $x^2+y^2-2x-4y+4=0$.
Hence, there are two such circles.

Answer:

Since, the circles are concentric.
$\Rightarrow$Centre of required circle = Centre of x² + y² − 4x − 6y − 3 = 0

The centre of the required circle is (2, 3).

We know that if a circle with centre (h, k) touches the y-axis, then h is the radius of the circle.
Thus, the radius is 2.

∴ Equation of the circle:
${\left(x-2\right)}^2+{\left(y-3\right)}^2=2^2$
 $\Rightarrow x^2+y^2-4x-6y+9=0$

Answer:

The general equation of the circle is x² + y² + 2gx + 2fy + c = 0
Now, it is passing through (0, 0)
∴ c = 0
Also, it is passing through (a, 0)
∴  a² + 2ag = 0
⇒ a(a + 2g) = 0
⇒a + 2g = 0
$\Rightarrow g=-\frac{a}{2}$
Again, it is passing through (0, b)
∴ b² + 2bf = 0
⇒ b(b + 2f) = 0
⇒b + 2f = 0
$\Rightarrow f=-\frac{b}{2}$
The coordinates of its centre are given by
$\left(-g, -f\right)=\left(\frac{a}{2}, \frac{b}{2}\right)$
 

Answer:

The general equation of the circle is x² + y² + 2gx + 2fy + c = 0 where the centre of the circle is (−g, −f)
Now, it is passing through (2, 3)
∴ 13 + 4g + 6f + c = 0                       .....(1)
Also, it is passing through (4, 5)
∴  41 + 8g + 10f + c = 0                  .....(2)
⇒g=−a2
Now, the centre lies on the straight line y − 4x + 3 = 0
∴ −f + 4g + 3 = 0                 .....(3)
⇒g=−a2
Solving (1), (2) and (3), we get
g = −2, f = −5 and c = 25
The equation of the circle is given by x² + y² − 4x − 10y + 25 = 0

Answer:

(2, −3) and (−2, 4) are the ends points of the diameter of a circle. The equation of this circle is $\left(x-2\right)\left(x+2\right)+\left(y+3\right)\left(y-4\right)=0$.

$$\begin{gathered} \Rightarrow x^2-4+y^2-4y+3y-12=0 \\ \Rightarrow x^2+y^2-y-16=0 ...\left(1\right) \end{gathered}$$

Equation (1) can be rewritten as
$$\begin{gathered} x^2+{\left(y-\frac{1}{2}\right)}^2-\frac{1}{4}-16=0 \\ \Rightarrow x^2+{\left(y-\frac{1}{2}\right)}^2=\frac{65}{4} \end{gathered}$$

∴ Centre is $\left(0,\frac{1}{2}\right)$ and radius is $\frac{\sqrt{65}}{2}$.

Answer:

Given:
$x^2+y^2+6x-14y-1=0$ ...(1)
And, $x^2+y^2-4x+10y-2=0$ ...(2)

Equations (1) and (2) can be rewritten as follows:
${\left(x+3\right)}^2+{\left(y-7\right)}^2=59$
And, ${\left(x-2\right)}^2+{\left(y+5\right)}^2=31$

Thus, the centres of the circles are (−3, 7) and (2, −5).

Hence, the equation of the circle, the end points of whose diameter are the centres of the given circles, is$\left(x+3\right)\left(x-2\right)+\left(y-7\right)\left(y+5\right)=0$, i.e. $x^2+y^2+x-2y-41=0$.

Answer:

According to the question:
Sides of the square are x = 6, x = 9, y = 3 and y = 6.

The vertices of the square are (6, 6), (9, 6), (9, 3) and (6, 3).
And, the vertices of two diagonals are (6, 6), (9, 3) and (9, 6), (6, 3).

Hence, the equation of the circle is $\left(x-6\right)\left(x-9\right)+\left(y-6\right)\left(y-3\right)$ or $x^2+y^2-15x-9y+72=0$.

Answer:

Given:
Sides of the rectangle:
x − 3y = 4     ...(1)
3x + y = 22   ...(2)
x − 3y = 14    ...(3)
And, 3x + y = 62    ...(4)

The intersection of (1) and (2) is (7, 1).
The intersection of (2) and (3) is (8, −2).
The intersection of (3) and (4) is (20, 2).
The intersection of (1) and (4) is (19, 5).

Hence, the vertices of the rectangle are (7, −1), (8, −2), (20, 2) and (19, 5).

The vertices of the diagonals are (7, −1), (20, 2) and (19, 5), (8, −2).

Thus, the required equation of the circle is $\left(x-7\right)\left(x-20\right)+\left(y-1\right)\left(y-2\right)=0$ or $x^2+y^2-27x-3y+142=0$.

Answer:

Putting x = 0 in 3x + 4y = 12:
y = 3

Putting y = 0 in 3x + 4y = 12:
x = 4

Thus, the line 3x + 4y = 12 meets the axes of coordinates at points A (0, 3) and B (4, 0).

The equation of the circle with AB as the diameter is $\left(x-0\right)\left(x-4\right)+\left(y-3\right)\left(y-0\right)=0$ or $x^2-4x+y^2-3y=0$.

Hence, the required equation is $x^2-4x+y^2-3y=0$.

Answer:

Case I:
If the required circle passes through the origin and (a, b), then the end points of the diameter of the circle will be (0, 0) and (a, b).
∴ Required equation of circle:
$\left(x-0\right)\left(x-a\right)+\left(y-0\right)\left(y-b\right)$
or $x^2+y^2-ax-by=0$

Case II:
If the required circle passes through the origin and (−a, −b), then the end points of the diameter of the circle will be (0, 0) and (−a, −b).
∴ Required equation of circle:
$\left(x-0\right)\left(x+a\right)+\left(y-0\right)\left(y+b\right)$
or $x^2+y^2+ax+by=0$


Hence, the equation of the required circle is $x^2+y^2\pm ax\pm by=0$.

Answer:

It is given that the end points of the diameter of the circle are (−4, 3) and (12, −1).
∴ Required equation of circle:
$\left(x+4\right)\left(x-12\right)+\left(y-3\right)\left(y+1\right)$
or $x^2+y^2-8x-2y-51=0$   ...(1)

Putting x = 0 in (1):
y² − 2y − 51 = 0
⇒ y² −  2y − 51 = 0
⇒ $y=1 \pm 2\sqrt{13}$

Hence, the intercepts made by it on the y-axis is $1+2\sqrt{13}-1+2\sqrt{13}=4\sqrt{13}$.

Answer:

Roots of equation x² + 2ax − b² = 0 are $-a\pm \sqrt{a^2+b^2}$.
Roots of equation x² + 2px − q² = 0 are $-p\pm \sqrt{p^2+q^2}$.

Therefore, coordinates of A and B are $\left(-a+\sqrt{a^2+b^2}, -p+\sqrt{p^2+q^2}\right) \mathrm{and} \left(-\mathrm{a}-\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}, -\mathrm{p}-\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}\right)$ respectively.

Hence, equation of circle is $\left(x+a-\sqrt{a^2+b^2}\right)\left(x+a+\sqrt{a^2+b^2}\right) + \left(y+p-\sqrt{p^2+q^2}\right)\left(y+p+\sqrt{p^2+q^2}\right)=0$.
$\Rightarrow {\left(x+a\right)}^2-a^2-b^2+{\left(y+p\right)}^2-p^2-q^2=0$
$\Rightarrow x^2+y^2+2ax+2yp-p^2-q^2=0$.

Also, radius of circle is $\sqrt{a^2+b^2+p^2+q^2}$.

Answer:

Given:
ABCD is a square with side a units.

Let AB and AD represent the x-axis and the y-axis, respectively.

Thus, the coordinates of B and D are (a, 0) and (0, a), respectively.

The end points of the diameter of the circle circumscribing the square are B and D.

Thus, equation of the circle circumscribing the square is $\left(x-a\right)\left(x-0\right)+\left(y-0\right)\left(y-a\right)=0$ or $x^2+y^2-a\left(x+y\right)=0$.

Answer:

The equation of the line can be rewritten as $x=\frac{y-6}{2}$.

Substituting the value of x in the equation of the circle, we get:

${\left(\frac{y-6}{2}\right)}^2+ y^2-2y-9=0$
$$\begin{gathered} \Rightarrow {\left(y-6\right)}^2+ 4y^2-8y-36=0 \\ \Rightarrow y^2+36-12y+ 4y^2-8y-36=0 \\ \Rightarrow 5y^2-20y=0 \\ \Rightarrow y^2-4y=0 \\ \Rightarrow y\left(y-4\right)=0 \\ \Rightarrow y=0,4 \end{gathered}$$

At y = 0, x = −3
At y = 4, x = −1

Therefore, the coordinates of A and B are $\left(-1,4\right) \mathrm{and} \left(-3,0\right)$.

∴ Equation of the circle with AB as its diameter:
$\left(x+1\right)\left(x+3\right)+\left(y-4\right)\left(y-0\right)=0$
$\Rightarrow x^2+4x+y^2-4y+3=0$