A) \[{{x}^{2}}+{{y}^{2}}-4x+10y-19=0\]
B) \[{{x}^{2}}+{{y}^{2}}+4x+10y-19=0\]
C) \[{{x}^{2}}+{{y}^{2}}+4x-10y-19=0\]
D) \[{{x}^{2}}+{{y}^{2}}-4x-10y+19=0\]
Correct Answer: D
Solution :
Clearly, \[{{m}_{CP}}\times {{m}_{AB}}=-1\] \[\Rightarrow \,\left( \frac{k-2}{h-3} \right)\times \left( \frac{k-8}{h-1} \right)=-1\] \[\therefore \] Locus of (h, k) is \[(x-1)(x-3)+(y-2)(y-8)=0\] i.e., \[{{x}^{2}}+{{y}^{2}}-4x-10y+19=0\]You need to login to perform this action.
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