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 :
Centre (2, ? 3) Circumference \[=10\pi \,\]Þ \[2\pi r=10\pi \]Þ \[r=5.\] From\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\], \[{{(x-2)}^{2}}+{{(y+3)}^{2}}={{5}^{2}}\] Þ \[{{x}^{2}}+{{y}^{2}}-4x+6y+13=25\] Þ \[{{x}^{2}}+{{y}^{2}}-4x+6y-12=0\], which is the required equation of the circle.
You need to login to perform this action.
You will be redirected in
3 sec
