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) \[{{x}^{2}}+{{y}^{2}}-4x-6y-12=0\]
E) \[{{x}^{2}}+{{y}^{2}}-4x-6y+12=0\]
Correct Answer: C
Solution :
\[\because \]Circumference of circle \[=10\pi \] \[\Rightarrow \] \[2\pi r=10\pi \] \[\Rightarrow \] \[r=5\] \[\therefore \]Equation of circle whose centre is\[(2,-3)\]and radius 5, is \[{{(x-2)}^{2}}+{{(y+3)}^{2}}={{5}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4x+6y+13=25\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4x+6y-12=0\]
You need to login to perform this action.
You will be redirected in
3 sec
