A) \[{{x}^{2}}+{{y}^{2}}+2x+2y+5=0\]
B) \[{{x}^{2}}+{{y}^{2}}-4x-6y+5=0\]
C) \[{{x}^{2}}+{{y}^{2}}-x-y+3=0\]
D) \[{{x}^{2}}+{{y}^{2}}+5x+2y=0\]
Correct Answer: B
Solution :
[b] Radius of the circle = Perpendicular Distance of (2, 3) from \[x+y=1\] is \[\frac{4}{\sqrt{2}}=2\sqrt{2}\] \[\therefore \] The required equation will be \[{{(x-2)}^{2}}+{{(y-3)}^{2}}=8\Rightarrow {{x}^{2}}+{{y}^{2}}-4x-6y+5=0\]
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