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 $S.$ Then point $(2,2)$ is

• A. lying on the circle $S$
• B. centre of circle $S$
• C. lying inside the circle $S$
• D. lying outside the circle $S$

Answer 1

The correct option is D lying outside the circle $S$

Required equation of the circle is,
$(x^2+y^2+13x-3y)+\lambda (11x+\frac{1}{2}y+\frac{25}{2})=0$
Putting point $(1,1)$,
$12+\lambda \left(24\right)=0\Rightarrow \lambda =-\frac{1}{2}$
Therefore, the equation of the circle is,
$(x^2+y^2+13x-3y)-\frac{1}{2}(11x+\frac{1}{2}y+\frac{25}{2})=0x^2+y^2+\frac{15}{2}x-\frac{13}{4}y-\frac{25}{4}=0$
Putting point $(2,2)$,
$4+4+15-\frac{13}{2}-\frac{25}{4}>0$

Hence, the point lies outside of the circle $S$.