A) \[{{x}^{2}}+{{y}^{2}}-3x-12y+2=0\]
B) \[{{x}^{2}}+{{y}^{2}}-3x+12y+2=0\]
C) \[{{x}^{2}}+{{y}^{2}}+3x+12y+2=0\]
D) None of these
Correct Answer: C
Solution :
Let centre be (h, k), then \[{{(h-3)}^{2}}+{{(k+2)}^{2}}={{(h+2)}^{2}}+{{k}^{2}}\] \[\Rightarrow 10h-4k-9=0\] Also the centre lies on the given line, so \[2h-k=3\]. On solving \[k=-6,\ h=-\frac{3}{2}\] Radius is\[{{(h-3)}^{2}}+{{(k+2)}^{2}}=\frac{145}{4}\], which is true for option (c) only.
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