JEE Main & Advanced
Sample Paper
JEE Main Sample Paper-23
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 \[(10{{r}^{2}})\,=10\left( \frac{1}{2} \right)=5\]