Question

The equation of the circle through the points of intersection of the circle $x^2+y^2-2x-4y+4=0$ and the line $x+2y=4$ and which touches the line $x+2y=0$, is

• A. $x^2+y^2-x-2y=0$
• B. $x^2+y^2-x+2y=0$
• C. $x^2+y^2+x-2y=0$
• D. None of these

Answer 1

Answer: (A)

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