question_answer

If the smallest radius of a circle passing through the intersection of \[{{x}^{2}}+{{y}^{2}}+2x=0\] and \[x-y=0\], is r then the value of \[(10\,{{r}^{2}})\] is equal to

A)  2                                

B)  4

C)  5                                

D)  10

Correct Answer: C

Solution :

On solving \[{{x}^{2}}+{{y}^{2}}\,+2x=0\] and \[x-y=0,\] we get A(0, 0) and B(-1, -1) Clearly, required circle is the circle described on AB as diameter. So radius \[=r=\frac{AB}{2}=\frac{\sqrt{2}}{2}\,=\frac{1}{\sqrt{2}}\] Hence \[(10{{r}^{2}})\,=10\left( \frac{1}{2} \right)=5\]