Question

The equation of the circle passing through $(1,2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$, is

• A. $x^2+y^2+6x-2y+9=0$
• B. $x^2+y^2-6x-2y+9=0$
• C. $x^2+y^2-6x-4y+9=0$
• D. $x^2+y^2-6x+4y+9=0$

Answer 1

Answer: (C)

The equation of the circle passing through the points of intersection of the given circles, is
$(x^2+y^2-8x-6y+21)$
$+\lambda \left(x^2+y^2-2x-15\right)=0\dots (i)$
If this circle passes through point $(1,2)$, then
$(1+4-8-12+21)+\lambda (1+4-2-15)=0$
$\Rightarrow 6\lambda =12$
$\Rightarrow \lambda =\frac{1}{2}$
On substituting $\lambda =\frac{1}{2}$ in Eq. (i), we get the equation of the required circle as
$x^2+y^2-8x-6y+21+\frac{x^2}{2}+\frac{y^2}{2}-x-\frac{15}{2}=0$
$\Rightarrow 2x^2+2y^2-16x-12y+42+x^2$
$+y^2-2x-15=0$
$\Rightarrow 3x^2+3y^2-18x-12y+27=0$
$\Rightarrow x^2+y^2-6x-4y+9=0$