Question

The equation of 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

Answer: (B)

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

Any circle passing through the points of intersection of the given line and circle has the equation

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

Its center $=(-\frac{\lambda }{2},-\frac{\lambda }{2})$

The circle is the smallest is $(-\frac{\lambda }{2},-\frac{\lambda }{2})$ is on the chord $x+y=1$

$\therefore -\frac{\lambda }{2},-\frac{\lambda }{2}=1\Rightarrow \lambda =-1.$

Putting this value for $\lambda ,$ the equation of the smallets circle is

$x^2+y^2-9-(x+y-1)=0\Rightarrow x^2+y^2-x-y-8=0$