A) \[{{x}^{2}}+{{y}^{2}}-6x+4y+25=0\]
B) \[{{x}^{2}}+{{y}^{2}}-6x+4y+12=0\]
C) \[{{x}^{2}}+{{y}^{2}}-6x+4y-12=0\]
D) \[{{x}^{2}}+{{y}^{2}}-6x+4y+13=0\]
E) \[{{x}^{2}}+{{y}^{2}}-6x+4y+9=0\]
Correct Answer: C
Solution :
Radius of circle = Perpendicular distance from \[(3,-2)\]to the line\[4x+3y+19=0\]and \[=\frac{4(3)+3(-2)+19}{\sqrt{16+9}}=5\] \[\therefore \]Required equation of circle is \[{{(x-3)}^{2}}+{{(y+2)}^{2}}={{5}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-6x+4y-12=0\]
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