question_answer

A chord AB drawn from the point \[A(0,3)\] on circle \[{{x}^{2}}+4x+{{(y-3)}^{2}}=0\]meets to M in such a way that \[AM=2AB,\]then the locus of point At will be

A) Straight line         

B) Circle

C) Parabola           

D) None of these           

Correct Answer: B

Solution :

Let \[M\,(h,k)\] Given, \[AM=2AB\] \[\Rightarrow \,\,AB+BM=2AB\] \[\Rightarrow \,\,AB=BM\] So B is mid point of AM \[B=\left( \frac{h}{2},\frac{k+3}{2} \right)\] \[\because \] Point-5 lies on the circle. \[\therefore \]  B satisfies the equation of circle, i.e., \[{{\left( \frac{h}{2} \right)}^{2}}+4\left( \frac{h}{2} \right)+{{\left( \frac{k+3}{2}-3 \right)}^{2}}=0\] \[\Rightarrow \,\frac{{{h}^{2}}}{4}+\frac{8h}{4}+\frac{{{(k-3)}^{2}}}{4}=0\] or \[{{x}^{2}}+{{y}^{2}}+8x-6y+9=0,\]which is a circle.