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. $x^2+y^2-4x-4y-3=0$
• B. $3(x^2+y^2)+4x-4y-3=0$
• C. $x^2+y^2-x-2y=0$
• D. $3(x^2+y^2)+4(x+y)-3=0$

Answer 1

The correct option is C

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

Family of circles is

$x^2+y^2-2x-4y+1+\lambda (x^2+y^2-1)=0$.

$(1+\lambda )x^2+(1+\lambda )y^2-2x-4y+(1-\lambda )=0$

$x^2+y^2-\frac{2}{1+\lambda }x-\frac{4}{1+\lambda }y+\frac{1-\lambda }{1+\lambda }=0$ ............(i)

Centre is $[\frac{1}{1+\lambda },\frac{2}{1+\lambda }]$

and radius = $\sqrt{{\left(\frac{1}{1+\lambda }\right)}^2+{\left(\frac{2}{1+\lambda }\right)}^2-\frac{1-\lambda }{1+\lambda }}=\frac{\sqrt{4+{\lambda }^2}}{1+\lambda }$.

Since it touches the line x + 2y = 0, hence

Radius = Perpendicular from centre to the line

i.e., $\left|\frac{\frac{1}{1+\lambda }+2\frac{2}{1+\lambda }}{\sqrt{1^2+2^2}}\right|$ = $\frac{\sqrt{4+{\lambda }^2}}{1+\lambda }$

$\Rightarrow \sqrt{5}=\sqrt{4+{\lambda }^2}\Rightarrow =\pm 1$

$\lambda =-1$ cannot be possible in case of circle. So $\lambda =1$.

Thus, from (i) $x^2+y^2-x-2y=0$ is the required equation of the circle.