A) \[{{x}^{2}}+{{y}^{2}}+4x+6y+12=0\]
B) \[{{x}^{2}}+{{y}^{2}}-4x+6y+12=0\]
C) \[{{x}^{2}}+{{y}^{2}}-4x+6y-12=0\]
D) \[{{x}^{2}}+{{y}^{2}}-4x-6y-12=0\]
Correct Answer: C
Solution :
Let \[r\] be the radius of a circle. \[\therefore \]Circumference of a circle\[=10\pi \] \[\Rightarrow \] \[2\pi r=10\pi \] \[\Rightarrow \] \[r=5\,\,unit\] \[\therefore \]Equation of circle whose centre is \[(2,\,\,-3)\] and radius \[5\], is \[{{(x-2)}^{2}}+{{(y+3)}^{3}}={{5}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4x+6y+4+9=25\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4x+6y-12=0\]
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