A) \[{{x}^{2}}+{{y}^{2}}-16x+4y-32=0\]
B) \[{{x}^{2}}+{{y}^{2}}+16x+4y-32=0\]
C) \[{{x}^{2}}+{{y}^{2}}+16x+4y+32=0\]
D) \[{{x}^{2}}+{{y}^{2}}+16x-4y+32=0\]
Correct Answer: A
Solution :
Here \[r=10\] (radius) Centre will be the point of intersection of the diameters, i.e. (8, ?2). Hence required equation is \[{{(x-8)}^{2}}+{{(y+2)}^{2}}={{10}^{2}}\Rightarrow {{x}^{2}}+{{y}^{2}}-16x+4y-32=0\].
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