The equation of the smallest circle passing through the intersection of the line \(x+y=1\) and the circle…
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
\(x^2+y^2-9-(x+y+1)=0\)
\(x^2+y^2-9-(x+y-1)=0\)
\(x^2+y^2-9-x+y-1=0\)
\(x^2+y^2-9+x+y-1=0\)
Solution
The family of circles passes through the intersection of circles \(x^2+y^2=9\) and line \(x+y=1\) is
\(\begin{aligned}
& \qquad\left(x^2+y^2-9\right)+\lambda(x+y-1)=0 \\
& \Rightarrow \quad x^2+y^2+\lambda x+\lambda y-(\lambda+9)=0 \text { having centre } \\
& \left(-\frac{\lambda}{2},-\frac{\lambda}{2}\right)
\end{aligned}\)
For the smallest circle, the line \(x+y=1\) must be diameter of the circle, so
\(-\frac{\lambda}{2}-\frac{\lambda}{2}=1 \Rightarrow \lambda=-1\)
So, equation of the required circle is
\(x^2+y^2-x-y-8=0\)
or, \(\left(x^2+y^2-9\right)-(x+y-1)=0\)