Question

The equation of the circle passing through $(1,1)$ and the points of intersection of $x^2+y^2+13x-3y=0$ and $11x+\frac{1}{2}y+\frac{25}{2}=0$ is

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

Answer 1

The correct option is D None of the above

The required equation of circle is
$(x^2+y^2+13x-3y)+\lambda (11x+\frac{1}{2}y+\frac{25}{2})=0$ .... (i)
This circle passes through $(1,1)$.
$(1^2+1^2+13-3)+\lambda (11+\frac{1}{2}+\frac{25}{2})=0$
Therefore, $12+\lambda \left(24\right)=0$
$\Rightarrow \lambda =-\frac{1}{2}$
On putting this value of $\lambda$ in Eq. (i), we get
$x^2+y^2+13x-3y-\frac{11}{2}x-\frac{1}{4}y-\frac{25}{4}=0$
$\Rightarrow 4x^2+4y^2+52x-12y-22x-y-25=0$
$\Rightarrow 4x^2+4y^2+30x-13y-25=0$.