Question

The equation of the circle having the center on the line $x+2y-3=0$ and passing through the points of intersection of the circles $x^2+y^2-2x-4y+1=0$ and $x^2+y^2-4x-2y+4=0$ is

• A. $x^2+y^2-6x+7=0$
• B. $x^2+y^2-6x-7=0$
• C. $x^2+y^2-2x-2y+1=0$
• D. $x^2+y^2+2x-4y+4=0$

Answer 1

Answer: (B)

$x^2+y^2-6x-7=0$

Circle through two given circles is $S_1+\lambda S_2=0$
$\Rightarrow (x^2+y^2-2x-4y+1)+\lambda (x^2+y^2-4x-2y+4)=0......\left(1\right)$
Its center is $(\frac{1+2\lambda }{1+\lambda },\frac{2+\lambda }{1+\lambda })$ which lies on the line $x+2y-3=0$
$\Rightarrow \frac{1+2\lambda +2(2+\lambda )}{1+\lambda }-3=0$
$\Rightarrow \lambda +2=0\Rightarrow \lambda =-2,$ substitute this in (1), we get
$(x^2+y^2-2x-4y+1)-2(x^2+y^2-4x-2y+4)=0$
$\Rightarrow x^2+y^2-6x-7=0.$
Hence, option 'A' is correct.