A) \[{{x}^{2}}+{{y}^{2}}+4x+4y-5=0\]
B) \[{{x}^{2}}+{{y}^{2}}-4x-4y-5=0\]
C) \[{{x}^{2}}+{{y}^{2}}-4x=13\]
D) \[{{x}^{2}}+{{y}^{2}}-4x-4y+5=0\]
Correct Answer: B
Solution :
Key Idea: Equation of circle with centre (h, k) and radius r is\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]. Since, circle passes through (4, 5) and centre is (2, 2). \[\therefore \]Radius \[=\sqrt{{{(4-2)}^{2}}+{{(5-2)}^{2}}}=\sqrt{4+9}=\sqrt{13}\] \[\therefore \]Equation of circle \[{{(x-2)}^{2}}+{{(y-2)}^{2}}={{(\sqrt{13})}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4x-4y+4+4-13=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4x-4y-5=0\]
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