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Q.
The equation of the circle passing through $(1, 2)$ and the points of intersection of the circles $x^{2}+y^{2}-8 x-6 \,y+21=0$ and $x^{2}+y^{2}-2 x-15=0$, is
TS EAMCET 2015
Solution:
The equation of the circle passing through the points of intersection of the given circles, is
$\left(x^{2}+y^{2}\right.-8 \,x-6 \,y+21) $
$+\lambda\left(x^{2}+y^{2}-2 \,x-15\right)=0 \ldots( i )$
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}-8 \,x-6\, y+21+\frac{x^{2}}{2}+\frac{y^{2}}{2}-x-\frac{15}{2}=0$
$ \Rightarrow 2 \,x^{2}+2 \,y^{2}-16\, x-12\, y+42+x^{2} $
$+y^{2}-2 \,x-15=0 $
$\Rightarrow 3 \,x^{2}+3 \,y^{2}-18\, x-12 \,y+27=0 $
$\Rightarrow x^{2}+y^{2}-6 \,x-4 \,y+9=0 $