Question

The equation of the smallest circle passing through the intersection of the line $x+y=1$ and the circle $x^2+y^2=9$ is

• A. $x^2+y^2+x+y-8=0$
• B. $x^2+y^2-x-y-8=0$
• C. $x^2+y^2-x+y-8=0$
• D. None of these

Answer 1

The correct option is B

$x^2+y^2-x-y-8=0$

Form the equation of the circle based on given information

Equation of the circle passing through the intersection of the circle $x^2+y^2=9$ and the line $x+y=1$is given as

$x^2+y^2-9+\lambda (x+y-1)=0$

$\Rightarrow$ $x^2+y^2+\lambda x+\lambda y-\left(9+\lambda \right)=0...\left(i\right)$

comparing with standard equation of circle $x^2+y^2+2gx+2fy+c=0$

$g=f=\frac{-\lambda }{2}$ , $c=-(\lambda +9)$

Radius is calculated as

$R=\sqrt{g^2+f^2-c}$

$=\sqrt{{\left(\frac{-\lambda }{2}\right)}^2+{\left(\frac{-\lambda }{2}\right)}^2+\left(\lambda +9\right)}$

$R=\frac{1}{\sqrt{2}}\sqrt{{(\lambda +1)}^2+17}$

The value of $R$ is minimum when $\lambda =-1$ as we need the smallest possible circle.

Substitute $\lambda =-1$ in $\left(i\right)$ we get

$x^2+y^2-x-y-8=0$

Hence, $x^2+y^2-x-y-8=0$ is the equation of the smallest circle intersecting the given line and given circle, so, option $\left(B\right)$ is the correct answer.