A) \[{{x}^{2}}+{{y}^{2}}-4x-6y-16=0\]
B) \[{{x}^{2}}+{{y}^{2}}-4x-6y-20=0\]
C) \[{{x}^{2}}+{{y}^{2}}-4x-6y-12=0\]
D) \[{{x}^{2}}+{{y}^{2}}+4x+6y-12=0\]
Correct Answer: C
Solution :
The intersection point of diameter lines is\[(2,\,\,3)\] which is the centre of circle. Now, radius\[=\sqrt{{{(5-2)}^{2}}+{{(7-3)}^{2}}}\] \[=\sqrt{9+16}=5\] \[\therefore \]Required equation of circle is \[{{(x-2)}^{2}}+{{(y-3)}^{2}}={{5}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4x-6y-12=0\]
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