Circle 5 Question 12
12. From the point $A(0,3)$ on the circle $x^{2}+4 x+(y-3)^{2}=0$, a chord $A B$ is drawn and extended to a point $M$ such that $A M=2 A B$. The equation of the locus of $M$ is… .
(1986, 2M)
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Solution:
- Given, $(x+2)^{2}+(y-3)^{2}=4$
Let the coordinate be $M(h, k)$, where $B$ is mid-point of $A$ and $M$.
$$ \Rightarrow \quad B \frac{h}{2}, \frac{k+3}{2} $$
But $A B$ is the chord of circle
$$ x^{2}+4 x+(y-3)^{2}=0 $$
Thus, $B$ must satisfy above equation.
$$ \begin{aligned} & \therefore \quad \frac{h^{2}}{4}+\frac{4 h}{2}+\frac{1}{2}(k+3)-3^{2}=0 \\ & \Rightarrow \quad h^{2}+k^{2}+8 h-6 k+9=0 \end{aligned} $$
$\therefore$ Locus of $M$ is the circle
$$ x^{2}+y^{2}+8 x-6 y+9=0 $$
