Question

The equation of the circle through the points of intersection of $x^2+y^2-1=0$, $x^2+y^2-2x-4y+1=0$ and touching the line $x+2y=0$, is

• A. $x^2+y^2+x+2y=0$
• B. $2(x^2+y^2)+x+2y=0$
• C. $x^2+y^2-x-2y=0$
• D. $2(x^2+y^2)-x-2y=0$

Answer 1

The correct option is C $x^2+y^2-x-2y=0$

Family of circles passing through intersection of given circles is $x^2+y^2-2x-4y+1+\lambda (x^2+y^2-1)=0$
$\Rightarrow (1+\lambda )x^2+(1+\lambda )y^2-2x-4y+(1-\lambda )=0$
Now, the above equation touches the lines $x+2y=0$.
So, number of intersection points $=1$
$\therefore 4(1+\lambda )y^2+(1+\lambda )y^2+4y-4y+1-\lambda =0\Rightarrow 5(1+\lambda )y^2+1-\lambda =0$

For the above equation to have only one root $\lambda =1$ $(\because \lambda \ne -1)$
Hence, the equation of the required circle is
$2x^2+2y^2-2x-4y=0\Rightarrow x^2+y^2-x-2y=0$