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Q.
If the smallest radius of a circle passing through the intersection of $x^{2}+y^{2}+2 x=0$ and $x-y=0$, is $r$ then the value of $\left(10 r^{2}\right)$ is equal to
Conic Sections
Solution:
On solving $x^{2}+y^{2}+2 x=0$ and $x-y=0$, we get
$A(0,0)$ and $B(-1,-1)$
Clearly, required circle is the circle described on $A B$ as diameter.
So radius $=r=\frac{A B}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$.
Hence $\left(10 r^{2}\right)=10\left(\frac{1}{2}\right)=5$
