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

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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}

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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

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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 

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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

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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

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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

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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

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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

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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}

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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

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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:

2021-02-14_13-59-26Case 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

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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

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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

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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

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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

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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.  

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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.

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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

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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:

2021-02-14_13-53-11Refer 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

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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.

Answer:

2021-02-14_13-52-54Refer to the adjoining diagram.

Let the center of the circle be ( t, 9-t)

\displaystyle \therefore r = \Big|  \frac{2t - 9 + t + 1}{\sqrt{2^1+1^1}}  \Big | = \Big| \frac{3t-8}{\sqrt{5}}  \Big |

\displaystyle \Rightarrow r^2 = \Big( \frac{3t-8}{\sqrt{5}} \Big )

Therefore the equation of the circle is:

(x-t)^2 + ( y -9+t)^2 = r^2

The circle passes through ( 2, 5) , therefore 

\displaystyle ( 2-t)^2 + ( 5-9+t)^2 = \Big( \frac{3t-8}{\sqrt{5}} \Big)^2

\Rightarrow 5 ( 2+t^2 - 4t + 16 + t^2 - 8t) = ( 9t^2 + 64 - 48 t)

\Rightarrow 10t^2 - 60t + 90= 9t^2 + 64- 48t

\Rightarrow t^2 - 12t + 36 = 0

\Rightarrow ( t-6)^2=0

\Rightarrow t = 6

Substituting t=6 , we get \displaystyle r = \frac{3\times 6 - 8}{\sqrt{5}} = \frac{10}{\sqrt{5}}

Therefore the equation of the circle is :   ( x -6)^2 + ( y-3)^2 = 20