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.You need to login to perform this action.
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