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Q.
The equation of the circle through the points of intersection of the circle $x^{2}+y^{2}-2 x-4 y+4=0$ and the line $x+2 y=4$ and which touches the line $x+2 y=0$, is
Conic Sections
Solution:
The equation of the circle passing through the points of intersection of the circle
$S = x ^{2}+ y ^{2}-2 x -4 y +4=0$
and the line $L = x +2 y -4=0$
is given by $S +\lambda L =0$
$\Rightarrow \left(x^{2}+y^{2}-2 x-4 y+4\right)+\lambda(x+2 y-4)=0$
or $ x^{2}+y^{2}+(\lambda-2) x+2(\lambda-2) y+4-4 \lambda=0$ .... (i)
The circle given by equation (i) touches the line $x + 2y = 0$.
Centre of the circle is $\left(-\frac{\lambda-2}{2},-(\lambda-2)\right)$.
Radius of the circle is
$\sqrt{\left(\frac{\lambda-2}{2}\right)^{2}+(\lambda-2)^{2}-4+4 \lambda}=\frac{1}{2} \sqrt{5 \lambda^{2}-4 \lambda+4}$
Hence, we have
$ \frac{\left|-\left(\frac{\lambda-2}{2}\right)-2(\lambda-2)\right| }{\sqrt{5}} = \frac{1}{2} \sqrt{5\lambda^2 - 4\lambda + 4}$
$\Rightarrow \sqrt{5} | \lambda - 2| = \sqrt{5\lambda^2 - 4\lambda +4}$
or $\lambda = 1$
So, the equation (i) becomes $x^{2}+y^{2}-x-2 y=0$
Which is the equation of the required circle.