question_answer

The locus of the midpoints of the chords drawn from point \[M(1,8)\]to the circle \[{{x}^{2}}+{{y}^{2}}-6x-4y-11=0\], is equal to

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., \[{{x}^{2}}+{{y}^{2}}-4x-10y+19=0\]