A) \[{{x}^{2}}-6x-10y+14=0\]
B) \[{{x}^{2}}-10x-6y+14=0\]
C) \[{{y}^{2}}-6x-10y+14=0\]
D) \[{{y}^{2}}-10x-6y+14=0\]
Correct Answer: D
Solution :
Let the centre be (h, k), then radius = h Also \[C{{C}_{1}}={{R}_{1}}+{{R}_{2}}\] or \[\sqrt{{{(h-3)}^{2}}+{{(k-3)}^{2}}}=h+\sqrt{9+9-14}\] \[\Rightarrow {{(h-3)}^{2}}+{{(k-3)}^{2}}={{h}^{2}}+4+4h\] \[\Rightarrow {{k}^{2}}-10h-6k+14=0\] or \[{{y}^{2}}-10x-6y+14=0\].
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