Question

The equation of the circle passing through $\left(1,1\right)$ and the points of intersection of $x^2+y^2+13x-3y=0$ and $2x^2+2y^2+4x-7y-25=0$ is

• A. $4x^2+4y^2+30x-10y-25=0$
• B. $4x^2+4y^2+30x-13y-25=0$
• C. $4x^2+4y^2-17x-10y+25=0$
• D. None of these

Answer 1

The correct option is B

$4x^2+4y^2+30x-13y-25=0$

Explanation for the correct answer:

Obtaining the equation of the circle:

Family of circle concept used:

Since the equation of a curve passing through the intersection of two circles, the equation is given by

$x^2+y^2+13x-3y+k\left(2x^2+2y^2+4x-7y-25\right)=0\dots \dots \left(1\right)$

Given that, this circle also passes through $\left(1,1\right)$.

By substituting these coordinates in the equation $\left(1\right)$ we get,

$\begin{array}{rcl}2+13-3+k\left(-24\right)& =& 0\\ k& =& \frac{-12}{-24}\\ k& =& \frac{1}{2}\end{array}$

Now, again substituting the value $k=\frac{1}{2}$ in the equation $\left(1\right)$ we get,

$x^2+y^2+13x-3y+\frac{1}{2}\left(2x^2+2y^2+4x-7y-25\right)=0$

$x^2+y^2+13x-3y+\left(x^2+y^2+2x-\frac{7}{2}y-\frac{25}{2}\right)=0$

$4x^2+4y^2+30x-13y-25=0$

Hence, option (B) is the correct answer.