Class 11: The Circle – Exercise 24.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: The general equation of a circle is $( x-a)^2 + ( y-b)^2 = r^2$ where $( a, b)$ is the center of the circle and $r$ is the radius of the circle.

Question 1: Find the equation of the circle with:

(i) Centre $(-2,3)$ and radius $4$ (ii) Centre $(a,b)$ and radius $\sqrt{a^2+b^2}$

(iii) Centre $(0, -1)$ and radius $1$ (iv) Centre $( a \cos \alpha , a \sin \alpha )$ and radius $a$

(v) Centre $(a, a)$ and radius $\sqrt{2} a$

Answer:

(i) Given: Centre $(-2,3)$ and radius $4$

Therefore the equation of the circle is:

$[ x - ( -2)]^2 + ( y - 3) ^2 = 4^2$

$\Rightarrow (x+2)^2 + ( y - 3)^2 = 4^2$

$\Rightarrow x^2 + 4 + 4x + y^2 +9 - 6y = 16$

$\Rightarrow x^2 + y^2 + 4x-6y - 3= 0$

(ii) Given: Centre $(a,b)$ and radius $\sqrt{a^2+b^2}$

Therefore the equation of the circle is:

$(x-a)^2 + ( y-b)^2 = ( \sqrt{a^2 + b^2})^2$

$\Rightarrow x^2 + a^2 - 2ax + y^2 + b^2 -2by = a^2 + b^2$

$\Rightarrow x^2 + y^2 - 2ax-2by=0$

(iii) Given: Centre $(0, -1)$ and radius $1$

Therefore the equation of the circle is:

$( x-0)^2 + [ y -(-1)]^2 = 1^2$

$\Rightarrow x^2 + ( y+1)^2 = 1$

$\Rightarrow x^2 + y^2 + 2y = 0$

(iv) Given: Centre $( a \cos \alpha , a \sin \alpha )$ and radius $a$

Therefore the equation of the circle is:

$( x - a \cos \alpha)^2 + ( y - a \sin \alpha)^2= a^2$

$\Rightarrow x^2 + a^2 \cos^2 \alpha - 2 x a \cos \alpha + y^2 + a^2 \sin^2 \alpha - 2ya \sin \alpha = a^2$

$\Rightarrow x^2 + y^2 + a^2 ( \sin^2 \alpha + \cos^2 \alpha) - 2xa \cos \alpha - 2ya \sin \alpha = a^2$

$\Rightarrow x^2 + y^2 - 2xa \cos \alpha - 2ya \sin \alpha = 0$

(v) Given: Centre $(a, a)$ and radius $\sqrt{2} a$

Therefore the equation of the circle is:

$(x-a)^2 + ( y-a)^2 = ( \sqrt{2} a)^2$

$\Rightarrow x^2 + a^2 - 2xa + y^2 + a^2 - 2ay = 2a^2$

$\Rightarrow x^2 + y^2 - 2ax - 2ay = 0$

$\\$

Question 2: Find the center and radius of each of the following circles:

i) $(x-1)^2 + y^2 = 4$ ii) $(x+5)^2 + ( y+1)^2= 9$

iii) $x^2 + y^2 - 4x + 6y = 5$ iv) $x^2 + y^2 - x + 2y - 3 = 0$

Answer:

i) Given equation $(x-1)^2 + y^2 = 4$

Comparing it with general equation of circle $( x-a)^2 + ( y-b)^2 = r^2$ where $( a, b)$ is the center of the circle and $r$ is the radius of the circle.

Center $= ( 1, 0)$ and Radius $= 2$

ii) Given equation $(x+5)^2 + ( y+1)^2= 9$

Comparing it with general equation of circle $( x-a)^2 + ( y-b)^2 = r^2$ where $( a, b)$ is the center of the circle and $r$ is the radius of the circle.

Center $= ( -5, -1)$ and Radius $= 3$

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

$\Rightarrow ( x-2)^2 + (y-3)^2 - 4 - 9 = 5$

$\Rightarrow (x-2)^2 + (y-3)^2 = 18$

Comparing it with general equation of circle $( x-a)^2 + ( y-b)^2 = r^2$ where $( a, b)$ is the center of the circle and $r$ is the radius of the circle.

Center $= ( 2, 3)$ and Radius $= 3\sqrt{2}$

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

$\Rightarrow ( x - \frac{1}{2})^2 + ( y + 1)^2- \frac{1}{4} - 1 - 3 = 0$

$\Rightarrow ( x - \frac{1}{2})^2 + ( y + 1)^2 = ( \frac{\sqrt{17}}{2} )^2$

Comparing it with general equation of circle $( x-a)^2 + ( y-b)^2 = r^2$ where $( a, b)$ is the center of the circle and $r$ is the radius of the circle.

Center $= ( \frac{1}{2} , -1)$ and Radius $= \frac{\sqrt{17}}{2}$

$\\$

Question 3: Find the equation of the circle whose center is $(1,2)$ and which passes through the point $(4,6)$ .

Answer:

Given circle whose center is $(1,2)$ and which passes through the point $(4,6)$

Radius $( r) = \sqrt{ ( 4-1)^2 + ( 6-2)^2} = \sqrt{9+16} = 5$

Therefore the equation of the circle is

$( x-1)^2 + ( y-2)^2 = 5^2$

$\Rightarrow x^2 + 1 - 2x + y^2 + 4 - 4y = 25$

$\Rightarrow x^2 + y^2 - 2x - 4y - 20 = 0$

$\\$

Question 4: Find the equation of the circle passing through the point of intersection of the lines $x+3y=0$ and $2x-7y=0$ and whose center is the point of intersection of the lines $x+y+1=0$ and $x-2y+4=0$ .

Answer:

The center of the circle is the point of intersection of lines $x+y+1=0$ and $x-2y+4=0$ which is $( -2, 1)$

The circle passes through the point of intersection of lines $x+3y=0$ and $2x-7y=0$ which is $(0,0)$

Therefore the Radius $= \sqrt{(-2-0)^2 + ( 1- 0)^2} = \sqrt{5}$

Therefore the equation of the circle is :

$[ x - ( -2)]^2 + ( y -1)^2 = (\sqrt{5})^2$

$\Rightarrow x^2 + 4 + 4x + y^2 + 1 - 2y = 5$

$\Rightarrow x^2 + y^2 + 4x - 2y = 0$

$\\$

Question 5: Find the equation of the circle whose center lies on the positive direction of y-axis at a distance $6$ from the origin and whose radius is $4$ .

Answer:

Given center $= ( 0, 6)$ and Radius $= 4$

Therefore the equation of the circle is :

$(x-0)^2 + ( y - 6)^2 = 4^2$

$\Rightarrow x^2 + y^2 + 36 - 12y = 16$

$\Rightarrow x^2 + y^2 - 12y + 20 = 0$

$\\$

Question 6: If the equations of two diameters of a circle are $2x+y=6$ and $3x+2y =4$ and the radius is $10$ , find the equation of the circle.

Answer:

Center of the circle is the point of intersection of lines $2x+y=6$ and $3x+2y =4$ which is $(8, -10)$

Therefore the equation of the circle is :

$(x-8)^2 + [y - ( -10)]^2 = 10^2$

$\Rightarrow x^2 + 64 - 16x + y^2 + 100 + 20 y = 100$

$\Rightarrow x^2 + y^2 - 16 x + 20 x + 64 = 0$

$\\$

Question 7: Find the equation of a circle

(i) which touches both the axes at a distance of $6$ units from the origin.

(ii) which touches x-axis at a distance $5$ from the origin and radius $6$ units

(iii) which touches both the axes and passes through the point $(2, 1)$

iv) passing through the origin , radius $17$ and ordinate of the center is $-15$

Answer:

Let $( h, k)$ be the center of the circle with radius $a$ . Thus the equation will be $(x-h)^2 + ( y - k)^2 = a^2$

(i) It is given that the circle passes through the points $( 6,0)$ and $( 0, 6)$

$\therefore ( 6-h)^2 + ( 0-k)^2 = 6^2$

$\Rightarrow 36 + h^2 - 12h + k^2 = 36$

$\Rightarrow h^2 + k^2 - 12h = 0$ … … … … … i)

Also $\therefore ( 0-h)^2 + ( 6-k)^2 = 6^2$

$\Rightarrow h^2 + 36 + k^2 - 12k = 36$

$\Rightarrow h^2 + k^2 - 12k = 0$ … … … … … ii)

From i) and ii) $12k = 12h \Rightarrow k = h$

From ii) we get , $2k^2 - 12k = 0$

$\Rightarrow k ( k-6) = 0$

$\Rightarrow k = 6$ , since $k > 0$

Therefore $h = 6$

Hence the equation of the circle is:

$( x - 6)^2 + ( y - 6)^2 = 36$

$\Rightarrow x^2 + y^2 - 12x - 12y + 36 = 0$

(ii) It is given that the circle with radius $6$ units touch the x-axis at a distance of $5$ units from origin.

Therefore center is $( 5, 6)$

Hence the equation of the required circle is:

$(x-5)^2 + ( y - 6)^2 = 6^2$

$\Rightarrow x^2 + 25 - 10x + y^2 + 36 - 12 y = 36$

$\Rightarrow x^2 + y^2 - 10x -12y + 25 = 0$

(iii) 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 $( 2, 1)$

Therefore $4 + 1 - 4a - 2a + a^2 = 0$

$\Rightarrow a^2 - 6a + 5 = 0$

$\Rightarrow a^2 - 5a - a + 5 = 0$

$\Rightarrow ( a-1)(a-5) = 0$

$a = 1 \text{ or } a = 5$

Hence the required equations are:

$x^2 + y^2 - 2x - 2y + 1 = 0$ or $x^2 + y^2 - 10x - 10y + 25 = 0$

iv) Given $k = -15, a = 17$

The equation passes through the point $( 0,0)$

Therefore the equation of the circle is:

$( 0 - h )^2 + ( 0 - 15 )^2 = 17^2$

$\Rightarrow h = \pm 8$

Hence the required equations of the circle are

$( x-8)^2 + ( y + 15)^2 = 17^2 \hspace{0.5cm} \Rightarrow x^2 + y^2 - 16x + 30 y = 0$

$( x+8)^2 + ( y + 15)^2 = 17^2 \hspace{0.5cm} \Rightarrow x^2 + y^2 + 16x + 30 y = 0$

$\\$

Question 8: Find the equation of the circle which has its center at the point $(3,4)$ and touches the straight line $5x+12y-1=0$ .

Answer:

Given the center of the circle $(3, 4)$

Perpendicular distance of $( 3, 4)$ from the line $5x + 12y = 1$ is

$\displaystyle d = \frac{5 \times 3 + 12 \times 4 - 1}{\sqrt{5^2 + 12^2}} = \frac{62}{13}$

Therefore the equation of the circle will be

$\displaystyle ( x - 3)^2 + ( y - 4)^2 = \Big( \frac{62}{13} \Big)^2$

$\Rightarrow 169 ( x^2 + 9 - 6x + y^2 + 16 - 8y) = 3844$

$\Rightarrow 169x^2 + 169y^2 - 1014 x - 1352 y + 381 = 0$

$\\$

Question 9: Find the equation of the circle which touches the axes and whose center lies on $x - 2y = 3$ .

Answer:

Let the circle touches $A( a, 0)$ and $B ( 0, a)$ on the axes. Therefore the center is $( a, a)$ and radius is $a$ .

Since the center is $(a, a)$ lies on $x - 2y = 3$ , we get

$a - 2a = 3 \hspace{0.5cm} \Rightarrow a = - 3$

Therefore the center is $( -3, -3)$ and radius $= 3$

Therefore the equation of the circle is

$[ x - ( -3)]^2 + [ y -(-3)]^2 = 3^2$

$\Rightarrow x^2 + y^2 + 6x + 6y + 9 = 0$

$\\$

Question 10: A circle whose center is the point of intersection of the lines $2x-3y+4=0$ and $3x + 4y -5 = 0$ passes through the origin. Find its equation.

Answer:

The intersection point of $2x-3y+4=0$ and $3x + 4y -5 = 0$ is given by $P \Big($ $\frac{-1}{17}$ $,$ $\frac{22}{17}$ $\Big)$

The circle passes through $( 0,0)$ , therefore

Radius $= \sqrt{ (\frac{-1}{17} - 0)^2 + ( \frac{22}{17} -0 )^2} = \sqrt{ \frac{485}{7}}$

Therefore the equation of the circle is:

$\Big[ x - \Big($ $\frac{-1}{17}$ $\Big) \Big]^2 + \Big[ y - \Big($ $\frac{22}{7}$ $\Big) \Big]^2 = \Big(\sqrt{ \frac{485}{7}} \Big)^2$

$\Big( x +$ $\frac{1}{17}$ $\Big)^2 + \Big( y -$ $\frac{22}{7}$ $\Big)^2 =$ $\frac{485}{7}$

$\\$

Question 11: A circle of radius $4$ units touches the coordinate axes in the first quadrant. Find the equations of its images with respect to the line mirrors $x = 0$ and $y = 0$ .

Answer:

Given circle has a radius of $4$ and touches the coordinate axes in first quadrant.

Therefore the center $= ( 4, 4)$

The image of $( 4, 4)$ on $y = 0$ is $( 4, -4)$ . Therefore the equation of the circle is

$( x - 4)^2 + [ y - ( -4)]^2 = 4^2$ $\Rightarrow x^2 + y^2 - 8x + 8y + 16 = 0$

The image of $( 4, 4)$ on $x = 0$ is $( -4, 4)$ . Therefore the equation of the circle is

$[x - (-4)]^2 + (y-4)^2 = 4^2$ $\Rightarrow x^2 + y^2 + 8x - 8y + 16 = 0$

$\\$

Question 12: Find the equations of the circles touching y-axis at $(0, 3)$ and making an intercept of $8$ units on the x-axis.

Answer:

Case 1: Center lies in 1st Quadrant

The circle touches y-axis at $L( 0,3)$ and makes an intercept $AB$ of $8$ units on x-axis

Therefore $AB = 8$

Let the required line be $( x-h)^2 + ( y - k)^2 = a^2$

In $\triangle CAM: CA^2 = CN^2 + AM^2$

$\Rightarrow CA^2 = 3^2 + 4^2$ $\Rightarrow CA^2 = 5$

Therefore $CL = CA = 5$

Therefore the coordinate of center $= ( 5, 3)$ and Radius $= 5$

Therefore equation of the circle is:

$( x-5)^2 + ( y - 3)^2 = 5^2$ $\Rightarrow x^2 + y^2 - 10x - 6y + 9 = 0$

Case 2: Center lies in 2nd Quadrant

Therefore the coordinate of center $= ( -5, 3)$ and Radius $= 5$

Therefore equation of the circle is:

$[ x-(-5)]^2 + ( y - 3)^2 = 5^2$ $\Rightarrow x^2 + y^2 + 10x - 6y + 9 = 0$

$\\$

Question 13: Find the equations of the circles passing through two points on y-axis at distances $3$ from the origin and having radius $5$ .

Answer:

The circle passes through $( 0, 3)$ and $( 0. -3)$ and Radius $= 5$

Let the center of the circle be $( h, k)$ . Therefore the equation of the circle is:

$( x-h)^2 + ( y -k)^2 = 5^2$ … … … … … i)

Substituting $(0,3)$ in equation i) we get

$( 0-h)^2 + ( 3-k)^2 = 25$

$\Rightarrow h^2 + k^2 - 6k = 16$ … … … … … ii)

Substituting $(0,-3)$ in equation i) we get

$( 0-h)^2 + ( -3-k)^2 = 25$

$\Rightarrow h^2 + k^2 + 6k = 16$ … … … … … iii)

Solving ii) and iii) we get $12k = 0 \hspace{0.5cm} \Rightarrow k = 0$

Therefore $h = \pm 4$

Hence the equation of the required circle is

$( x \pm 4)^2 + y^2 = 25 \hspace{0.5cm} \Rightarrow x^2 + y^2 \pm 8x - 9 = 0$

$\\$

Question 14: If the lines $2x - 3y =5$ and $3x -4y =7$ are the diameters of a circle of area $154$ square units, then obtain the equation of the circle.

Answer:

The point of intersection of $2x - 3y =5$ and $3x -4y =7$ is $( 1, -1)$

Therefore center $= ( 1, -1)$

Given Area of the circle $= 154$ sq. units

$\Rightarrow \pi r^2 = 154$

$\displaystyle \Rightarrow r^2 = \frac{7}{22} \times 154 = 49$

$\therefore r = 7$

Therefore the equation of the circle is

$( x - 1)^2 + [ y - ( -1)]^2 = 7^2$

$\Rightarrow x^2 + y^2 - 2x + 2y - 47 = 0$

$\\$

Question 15: If the line $y = \sqrt{3} x + k$ touches the circle $x^2 + y^2 = 16$ , then find the value of $k$ .

Answer:

The center of the circle $x^2 + y^2 = 16$ is $( 0,0)$ and Radius $= 4$

Perpendicular distance from the center to the tangent $y = \sqrt{3} + k$ is equal to the radius

$\displaystyle \therefore \Bigg| \frac{ \sqrt{3}(0) - 0 + k}{\sqrt{ (\sqrt{3})^2+ 1^2}} \Bigg| = 4$

$\displaystyle \Rightarrow \frac{k}{2} = \pm 4$

$\Rightarrow k = \pm 8$

$\\$

Question 16: Find the equation of the circle having $(1,-2)$ as its center and passing through the intersection of the lines $3x+y=14$ and $2x+5y = 18$ .

Answer:

Point of intersection of $3x+y=14$ and $2x+5y = 18$ is $( 4, 2)$ .

Center of the circle is $( 1, -2)$ .

Therefore the radius $= \sqrt{(1-4)^2 + ( -2-2)^2} = \sqrt{25} = 5$

Therefore the equation of the circle is:

$( x-1)^2 + [ y - (-2)]^2 = 5^2$ $\Rightarrow x^2 + y^2 = 2x + 4y -20=0$

$\\$

Question 17: If the lines $3x-4y+4=0$ and $6x-8y-7 =0$ are tangents to a circle, then find the radius of the circle.

Answer:

Given lines:

$\displaystyle 3x-4y+4=0 \hspace{0.5cm} \Rightarrow \text{ Slope } = \frac{3}{4}$

$\displaystyle 6x-8y-7 =0 \hspace{0.5cm} \Rightarrow \text{ Slope } = \frac{6}{8} = \frac{3}{4}$

Therefore the two lines are parallel to each other. The distance between the two lines is

$\displaystyle d = \Bigg| \frac{c_1 - c_2 }{\sqrt{a^2+b^2}} \Bigg | = \Bigg| \frac{4 + \frac{7}{2}}{\sqrt{3^2 + 4^2}} \Bigg | = \frac{15}{10} = \frac{3}{2}$

Therefore the radius of the circle is $\displaystyle \frac{3}{4}$ units.

$\\$

Question 18: Show that the point $(x, y)$ given by $\displaystyle x = \frac{2at}{1+t^2}$ and $\displaystyle y = a \Big( \frac{1-t^2}{1+t^2} \Big)$ lies on a circle for all real values of $t$ such that $-1 \leq t \leq 1$ , where $a$ is any given real number.

Answer:

Given: $\displaystyle x = \frac{2at}{1+t^2}$ and $\displaystyle y = a \Big( \frac{1-t^2}{1+t^2} \Big)$

Squaring and adding we get:

$\displaystyle x^2 + y^2 = \frac{4a^2t^2}{(1+t^2)^2} + a^2 \Big( \frac{(1-t^2)^2}{(1+t^2)^2} \Big)$

$\displaystyle \Rightarrow x^2 + y^2 = \frac{4a^2t^2 + a^2 ( 1 + t^4 - 2t^2)}{( 1+t^2)^2}$

$\displaystyle \Rightarrow x^2 + y^2 = \frac{4a^2t^2 + a^2 + a^2t^4 - 2a^2t^2}{( 1+t^2)^2}$

$\displaystyle \Rightarrow x^2 + y^2 = \frac{2a^2t^2 + a^2t^4 + a^2}{( 1+t^2)^2}$

$\displaystyle \Rightarrow x^2 + y^2 = \frac{a^2( 1+t^2)^2}{( 1+t^2)^2}$

$\displaystyle \Rightarrow x^2 + y^2 = a^2$

Hence the above equation represents the equation of the circle on which $(x, y)$ lies.

$\\$

Question 19: The circle $x^2 + y^2- 2x - 2y + 1 = 0$ is rolled along the positive direction of x-axis and makes one complete roll. Find its equation in new-position.

Answer:

The equation of the circle is:

$x^2 + y^2- 2x - 2y + 1 = 0$ $\Rightarrow (x-1)^2 + ( y -1)^2 =1$

Therefore center is $( 1,1)$ and Radius $= 1$

Distance covered in one roll $= 2\pi (1) = 2\pi$ .

Therefore the center would move $2\pi$ towards right direction.

Therefore the new center would be $( 1 + 2\pi , 1)$

Hence the new equation of the circle is: $( x - 1 - 2\pi)^2 + ( y -1)^2 = 1$

$\\$

Question 20: One diameter of the circle circumscribing the rectangle $ABCD$ is $4y=x+7$ . If the coordinates of $A$ and $B$ are $(- 3, 4)$ and $(5, 4)$ respectively, find the equation of the circle.

Answer:

Refer to the adjoining diagram.

Given $A( -3, 4)$ and $B ( 5, 4)$

Slope of $\displaystyle AB = \frac{4-4}{5-(-3)} = 0$

Mid point of $AB = ( 1, 4)$

The equation of the perpendicular bisector $x=1$

The point of intersection of $4y = x + 7$ and $x = 1$ is $( 1,2)$

Therefore the center is $( 1,2 )$

Radius $= \sqrt{ ( 1 -(-3))^2 + (2 - 4)^2} = \sqrt{4^2 + 2^2} = \sqrt{20}$

Therefore the equation of the circle is:

$( x-1)^2 + ( y -2)^2 = ( \sqrt{20})^2$ $\Rightarrow x^2 + y^2 - 2x - 4y - 15 = 0$

$\\$

Question 21: If the line $2x - y + 1 = 0$ touches the circle at the point $(2, 5)$ and the center of the circle lies on the line $x + y - 9 = 0$ . Find the equation of the circle.