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 \[(10{{r}^{2}})\,=10\left( \frac{1}{2} \right)=5\]You need to login to perform this action.
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