Class 11: The Circle – Exercise 24.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: If the end points of the diameter is $(x_1, y_1)$ and $( x_2, y_2)$ , then the equation of the circle is given by: $(x-x_1)(x-x_2) + ( y - y_1) ( y - y_2) = 0$

Question 1: Find the equation of the circle, the end points of whose diameter are $(2, - 3)$ and $(-2,4)$ ; Find its center and radius.

Answer:

Given the end points of the diameter as $( 2, -3) \ \ \& \ \ ( -2, 4)$ .

Therefore the equation of circle is :

$(x-2)[x-( -2)] + [ y - (-3)] ( y - 4) = 0$

$\Rightarrow x^2 - 4 + y^2 - y - 12= 0$

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

$\displaystyle \therefore g = 0, \hspace{0.5cm} f = \frac{-1}{2}$

$\displaystyle \therefore \text{ Center } = ( -g, -f) = \Big( 0, \frac{1}{2} \Big)$

$\displaystyle \text{ Radius }= \sqrt{g^2 + f^2 - c} = \sqrt{(0)^2 + ( \frac{-1}{2})^2 - ( -16) } = \frac{\sqrt{65}}{2}$

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Question 2: Find the equation of the circle the end points of whose diameter are the centers of the circles $x^2 + y^2 +6x-14y -1 = 0$ and $x^2 + y^2 -4x+10y -2 = 0$ .

Answer:

Given equation of the circle 1:

$x^2 + y^2 +6x-14y -1 = 0$ $\Rightarrow \text{ Center } = ( -g, -f) = ( -3, 7)$

Given equation of the circle 2:

$x^2 + y^2 -4x+10y -2 = 0$ $\Rightarrow \text{ Center } = ( -g, -f) = ( 2, -5)$

Therefore the equation of the required circle is:

$[x - ( -3) ] (x-2) + ( y - 7)[ y - ( -5) ] = 0$

$\Rightarrow x^2 + 3x - 2x - 6 + y^2 - 7y + 5y - 35 = 0$

$\Rightarrow x^2 + y^2 + x - 2y - 41 = 0$

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Question 3: The sides of a square are $x= 6, x = 9, y = 3$ md $y = 6$ . Find the equation of a circle drawn on the diagonal of the square as its diameter.

Answer:

Refer to the adjoining figure.

The coordinates are $A( 6, 3), B( 9, 3), C( 9, 6) , D(6,6)$

Therefore the equation of the circle is with $AC$ as the diameter is:

$( x-6)(x-9) + ( y-3)(y-6) = 0$

$\Rightarrow x^2 - 15x + 54 + y^2 -9y +18=0$

$\Rightarrow x^2 + y^2 - 15x - 9y + 72 = 0$

Therefore the equation of the circle is with $BD$ as the diameter is:

$( x-6)(x-6) + ( y-3)(y-6) = 0$

$\Rightarrow x^2 - 15x + 54 + y^2 -9y +18=0$

$\Rightarrow x^2 + y^2 - 15x - 9y + 72 = 0$

These are the same circle also as you can see from the equations.

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Question 4: Find the equation of the circle circumscribing the rectangle whose sides are $x - 3y = 4, 3x + y =22, x - 3y =14$ and $3x + y = 62$ .

Answer:

Given equations:

$x - 3y = 4$ … … … … … i) $3x + y =22$ … … … … … ii)

$x - 3y =14$ … … … … … iii) $3x + y = 62$ … … … … … iv)

Solving i) & ii) , ii) & iii), iii) & iv) and iv) & i) we get $A( 7, 1), B( 8, -2), C( 20, 2) , D(19, 5)$

Therefore the equation of the circle is with $AC$ as the diameter is:

$( x-7)(x-20) + ( y-1)(y-2) = 0$

$\Rightarrow x^2 - 27x + 140 + y^2 -3y +2=0$

$\Rightarrow x^2 + y^2 - 27x - 3y + 142 = 0$

Therefore the equation of the circle is with $BD$ as the diameter is:

$( x-8)(x-19) + ( y-(-2))(y-5) = 0$

$\Rightarrow x^2 - 27x + 152 + y^2 -3y -10=0$

$\Rightarrow x^2 + y^2 - 27x - 3y + 142 = 0$

These are the same circle also as you can see from the equations.

$\\$

Question 5: Find the equation of the circle passing through the origin and the points where the line $3x + 4y = 12$ meets the axes of coordinates.

Answer:

Given line $3x + 4y = 12$ .

The intercepts on the axes are $A(0,3) \text{ and } B( 4, 0)$

Since the circle passes through origin, A and B, we can say that AB is the diameter of the required circle.

Therefore the equation of the circle is with $AB$ as the diameter is:

$( x-0)(x-4) + ( y-3)(y-0) = 0$

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

$\\$

Question 6: Find the equation of the circle which passes through the origin and cuts off intercepts $a$ and $b$ respectively from x and y-axes.

Answer:

Case 1: Given that the circle passes through $(0,0), A(0, b)$ and $B ( a, 0)$ .

Therefore the equation of the circle is with $AB$ as the diameter is:

$( x-a)(x-0) + ( y-0)(y-b) = 0$

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

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

Case 1: Given that the circle passes through $(0,0), A(0, -b)$ and $B ( -a, 0)$ .

Therefore the equation of the circle is with $AB$ as the diameter is:

$( x-(-a))(x-0) + ( y-0)(y-(-b)) = 0$

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

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

$\\$

Question 7: Find the equation of the circle whose diameter is the line segment joining $(- 4, 3)$ and $(12, -1)$ . Find also the intercept made by it on y-axis.

Answer:

Given the end points of the diameter as $( -4, 3) \ \ \& \ \ ( 12, -1)$ .

Therefore the equation of circle is :

$[x-( -4)] (x-12) + ( y - 3)[ y - (-17)] = 0$

$\Rightarrow (x+4)(x-12)+(y-3)(y+1)= 0$

$\Rightarrow x^2 + y^2 - 8x -2y -51 = 0$ … … … … … i)

To find y intercepts, put $x = 0$ in i)

Therefore $y^2 - 2y - 51 = 0$

$\Rightarrow y = 1 \pm 2\sqrt{13}$

Therefore the intercept points are $( 0, 1 + 2\sqrt{13}) , ( 0, -2\sqrt{13})$

Hence the intercept between the y-axis is

$= \sqrt{ (0-0)^2 + ( 1 - 2\sqrt{13} - 1 - 2\sqrt{13})^2 } = \sqrt{(-4\sqrt{13})^2} = 4\sqrt{13}$ units

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Question 8: The abscissae of the two points $A$ and $B$ are the roots of the equation $x^2 + 2ax - b^2 = 0$ and their ordinates are the roots of the equation $x^2 + 2px - q^2 =0$ . Find the equation of the circle with $AB$ as diameter. Also, find its radius.

Answer:

Roots of equation $x^2 + 2ax - b^2 = 0$ are $-a \pm \sqrt{a^2 + b^2}$

Roots of equation $x^2 + 2px - q^2 =0$ are $-p \pm \sqrt{p^2 + q^2}$

Therefore the coordinates of A and B are $( -a + \sqrt{a^2 + b^2} ,-p + \sqrt{p^2 + q^2})$ and $( -a - \sqrt{a^2 + b^2} ,-p - \sqrt{p^2 + q^2})$

Therefore the equation of the circle is:

$(x + a - \sqrt{a^2 + b^2}) ( x + a + \sqrt{a^2 + b^2}) + (y + p - \sqrt{p^2 + q^2}) ( y +p + \sqrt{p^2 + q^2}) = 0$

$\Rightarrow ( x+a)^2 - a^2 - b^2 + ( y+p)^2 - p^2 - q^2 = 0$

$\Rightarrow x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0$

Radius $= \sqrt{g^2+f^2-c} = \sqrt{ a^2 + p^2 + ( b^2 + q^2) } = \sqrt{a^2 + b^2 + p^2 + q^2}$

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Question 9: $ABCD$ is a square whose side is $a$ ; taking $AB$ and $AD$ as axes, prove that the equation of the circle circumscribing the square is $x^2 + y^2 - a( x+y) = 0$ .

Answer:

Since $AB$ and Ad are x-axis and y-axis respectively, the coordinates are $A(0, 0), B(a, 0), C(a, a), D( 0, a)$

Since $BD$ is a diameter, therefore equation of circle is:

$(x-a)(x-0) + ( y - 0) ( y-a) = 0$

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

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

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

Also $AC$ is a diameter, therefore equation of circle is:

$(x-0)(x-a) + ( y - 0) ( y-a) = 0$

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

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

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

Both are the same circles.

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Question 10: The line $2x - y + 5 = 0$ meets the circle $x^2 + y^2 - 2 y - 9 = 0$ at $A$ and $B$ . Find the equation of the circle on $AB$ as diameter.

Answer:

Given equations:

$2x - y + 5 = 0$ … … … … … i) $x^2 + y^2 - 2 y - 9 = 0$ … … … … … ii)

Solving i) and ii) simultaneously,

$x^2 + ( 2x+6)^2 - 2 ( 2x+6) - 9 = 0$

$\Rightarrow x^2 + 4x^2 + 36 + 24 x - 4x - 12 - 9 = 0$

$\Rightarrow 5x^2 + 20 x + 15=0$

$\Rightarrow x^2 + 4x + 3 = 0$

$\Rightarrow ( x+3) ( x+1) = 0$

$\Rightarrow x= - 3 \text{ or } x = - 1$

$\therefore y = 0 \text{ or } y = 4$

$\therefore A( -3, 0) \text{ and } B ( -1, 4)$

Therefore equation of the circle is:

$[ x - ( -3)][x-(-1)] + ( y-0)(y-4) = 0$

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

$\\$

Question 11: Find the equation of the circle which circumscribes the triangle formed by the lines $x =0, y = 0$ and $lx +my =1$ .

Answer:

Given equations:

$x =0$ … … … … … i) $y = 0$ … … … … … ii) $lx +my =1$ … … … … … iii)

Solving i), ii) and iii) we get the vertices as

$\displaystyle C ( 0,0), A \Big( 0, \frac{1}{m} \Big), B \Big( \frac{1}{l} , 0 \Big)$

Therefore the equation of the circle with $AB$ as the diameter is:

$\displaystyle ( x - 0) \Big( x - \frac{1}{l} \Big) + \Big( y - \frac{1}{m} \Big) ( y - 0) = 0$

$\displaystyle \Rightarrow x^2 - \frac{x}{l} + y^2 - \frac{y}{m} = 0$

$\\$

Question 12: Find the equations of the circles which pass through the origin and cut off equal chords of $\sqrt{2}$ units from the lines $y =x$ and $y = - x$ .