Question

The equation of the circle passing through points of intersection of the circle $x^2+y^2-2x-4y+4=0$ and the line $x+2y=4$ and 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. $x^2+y^2-x+2y=0$

Answer 1

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

Equation of any circle passing through points of intersection of the given circle and the line is
$(x^2+y^2-2x-4y+4)+\lambda (x+2y-4)=0$
$\Rightarrow x^2+y^2+(\lambda -2)x+(2\lambda -4)y+4(1-\lambda )=0\cdots \left(1\right)$

It will touch the line $x+2y=0$ if solution of equation $\left(1\right)$ and $x+2y=0$ is unique.

Hence the roots of the equation
$(-2y{)}^2+y^2+(\lambda -2\left)\right(-2y)+(2\lambda -4)y+4(1-\lambda )=0$
$\Rightarrow 5y^2+4(1-\lambda )=0$ must be equal.
$0-4\times 5\times 4(1-\lambda )=0$
$\Rightarrow \lambda =1$
From $\left(1\right)$, the required circle is $x^2+y^2-x-2y=0$