Question

Equation of the circle passing through the origin and through the points of intersection of the circle $x^2+y^2-2x+4y-20=0$ and the line $x+y-1=0$ is

• A. $x^2+y^2-20x+15y=0$
• B. $x^2+y^2+33x+33y=0$
• C. $x^2+y^2-22x-16y=0$
• D. $2x^2+2y^2-4x-5y=0$

Answer 1

The correct option is A $x^2+y^2-20x+15y=0$

As given in the problem, we have to find point of intersection of circle and a line.

Equation of circle is given by,

$x^2+y^2-2x+4y-20=0$ Equation (1)

Equation of line is given by,

$x+y-1=0$

$\therefore y=1-x$

Put this value in equation (1), we get,

$x^2+{(1-x)}^2-2x+4(1-x)-20=0$

$\therefore x^2+1-2x+x^2-2x+4-4x-20=0$

$\therefore 2x^2-8x-15=0$

$\therefore x=\frac{-(-8)\pm \sqrt{{(-8)}^2-4\times 2\times (-15)}}{2\times 2}$

$\therefore x=\frac{8\pm \sqrt{64+120}}{4}$

$\therefore x=\frac{8\pm \sqrt{184}}{4}$

$\therefore x=\frac{8\pm \sqrt{4\times 46}}{4}$

$\therefore x=\frac{8\pm 2\sqrt{46}}{4}$

$\therefore x=\frac{4\pm \sqrt{46}}{2}$

$\therefore x=\frac{4+\sqrt{46}}{2}$ and $\therefore x=\frac{4-\sqrt{46}}{2}$

When $x=\frac{4+\sqrt{46}}{2}$, $y=1-\frac{4+\sqrt{46}}{2}$

$\therefore y=\frac{2-(4+\sqrt{46})}{2}$

$\therefore y=\frac{-2-\sqrt{46}}{2}$

When $x=\frac{4-\sqrt{46}}{2}$, $y=1-\frac{4-\sqrt{46}}{2}$

$\therefore y=\frac{2-(4-\sqrt{46})}{2}$

$\therefore y=\frac{-2+\sqrt{46}}{2}$

Thus, Points of intersection of circle with line are $A(\frac{4+\sqrt{46}}{2},\frac{-2-\sqrt{46}}{2})$ and $B(\frac{4-\sqrt{46}}{2},\frac{-2+\sqrt{46}}{2})$

Now as given in the problem, we have to find equation of circle passing through these points and origin $C(0,0)$

Let equation of this circle is $x^2+y^2+2gx+2fy+c=0$

As point A lies on this circle, it must satisfy equation of circle.

$\therefore {\left(\frac{4+\sqrt{46}}{2}\right)}^2+{\left(\frac{-2-\sqrt{46}}{2}\right)}^2+2g\left(\frac{4+\sqrt{46}}{2}\right)+2f\left(\frac{-2-\sqrt{46}}{2}\right)+c=0$

$\therefore \frac{16+8\sqrt{46}+46}{4}+\frac{4+4\sqrt{46}+46}{4}+g(4+\sqrt{46})+f(-2-\sqrt{46})+c=0$

$\therefore \frac{20+12\sqrt{46}+92}{4}+g(4+\sqrt{46})-f(2+\sqrt{46})+c=0$

$\therefore 5+3\sqrt{46}+23+g(4+\sqrt{46})-f(2+\sqrt{46})+c=0$

$\therefore c=-5-3\sqrt{46}-23-g(4+\sqrt{46})+f(2+\sqrt{46})$ Equation (1)

Also, point B lies on circle and must satisfy equation of circle

$\therefore {\left(\frac{4-\sqrt{46}}{2}\right)}^2+{\left(\frac{-2+\sqrt{46}}{2}\right)}^2+2g\left(\frac{4-\sqrt{46}}{2}\right)+2f\left(\frac{-2+\sqrt{46}}{2}\right)+c=0$

$\therefore \frac{16-8\sqrt{46}+46}{4}+\frac{4-4\sqrt{46}+46}{4}+g(4-\sqrt{46})+f(-2+\sqrt{46})+c=0$

$\therefore \frac{20-12\sqrt{46}+92}{4}+g(4-\sqrt{46})+f(-2+\sqrt{46})+c=0$

$\therefore 5-3\sqrt{46}+23+g(4-\sqrt{46})+f(-2+\sqrt{46})+c=0$

$\therefore c=-5+3\sqrt{46}-23-g(4-\sqrt{46})-f(-2+\sqrt{46})$ Equation (2)

From equation (1) an (2), we get,

$5-3\sqrt{46}-23-g(4+\sqrt{46})+f(2+\sqrt{46})=5+3\sqrt{46}-23-g(4-\sqrt{46})-f(-2+\sqrt{46})$

$\therefore -3\sqrt{46}-4g-g\sqrt{46}+2f+f\sqrt{46}=3\sqrt{46}-4g+g\sqrt{46}+2f-f\sqrt{46}$

$\therefore 2g\sqrt{46}-2f\sqrt{46}=-6\sqrt{46}$

$\therefore g-f=-3$

$\therefore g=f-3$ Equation (3)

Similarly, origin $O(0,0)$ also lies on the same circle. Thus, it must satisfy the equation of circle.

$\therefore {\left(0\right)}^2+{\left(0\right)}^2+2g\left(0\right)+2f\left(0\right)+c=0$

$\therefore c=0$

Thus, from equation (1),

$-5-3\sqrt{46}-23-(f-3)(4+\sqrt{46})+f(2+\sqrt{46})=0$

$\therefore -5-3\sqrt{46}-23-(4f+\sqrt{46}f-12-3\sqrt{46})+2f+f\sqrt{46}=0$

$\therefore -5-3\sqrt{46}-23-4f-\sqrt{46}f+12+3\sqrt{46}+2f+f\sqrt{46}=0$

$\therefore 2f=-16$

$\therefore f=-8$

Put this value in equation (3), we get,

$g=-8-3$

$g=-11$

Thus, equation of circle is given by,

$x^2+y^2+2(-11)x+2(-8)y=0$

$\therefore x^2+y^2-22x-16y=0$

Thus, answer is option (C)