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A chord $A B$ is drawn from the point $A(0,3)$ on
the circle $x^{2}+4 x+(y-3)^{2}=0,$ and is extended to $M$ such that $A M=2 A B$. The locus
of $M$ is
Options:
the circle $x^{2}+4 x+(y-3)^{2}=0,$ and is extended to $M$ such that $A M=2 A B$. The locus
of $M$ is
Solution:
2908 Upvotes
Verified Answer
The correct answer is:
$x^{2}+y^{2}+8 x-6 y+9=0$
Given, $A M=2 A B$
$\Rightarrow B$ is mid-point of $A M$
$\therefore$ Coordinate of $B$ is $\left(\frac{0+x_{1}}{2}, \frac{3+y_{1}}{2}\right)$
$$
=\left(\frac{x_{1}}{2}, \frac{y_{1}+3}{2}\right)
$$
since, $B$ lies on the circle $x^{2}+4 x+(y-3)^{2}=0$
$\therefore\left(\frac{x_{1}}{2}\right)^{2}+4\left(\frac{x_{1}}{2}\right)+\left(\frac{y_{1}+3}{2}-3\right)^{2}=0$
$\Rightarrow \quad \frac{x_{1}^{2}}{4}+2 x_{1}+\left(\frac{y_{1}-3}{2}\right)^{2}=0$
$\Rightarrow \quad \frac{x_{1}^{2}}{4}+2 x_{1}+\frac{y_{1}^{2}+9-6 y_{1}}{4}=0$
$\Rightarrow \quad x_{1}^{2}+y_{1}^{2}+8 x_{1}-6 y_{1}+9=0$
Hence, locus of a point is $x^{2}+y^{2}+8 x-6 y+9=0$
$\Rightarrow B$ is mid-point of $A M$
$\therefore$ Coordinate of $B$ is $\left(\frac{0+x_{1}}{2}, \frac{3+y_{1}}{2}\right)$
$$
=\left(\frac{x_{1}}{2}, \frac{y_{1}+3}{2}\right)
$$
since, $B$ lies on the circle $x^{2}+4 x+(y-3)^{2}=0$
$\therefore\left(\frac{x_{1}}{2}\right)^{2}+4\left(\frac{x_{1}}{2}\right)+\left(\frac{y_{1}+3}{2}-3\right)^{2}=0$
$\Rightarrow \quad \frac{x_{1}^{2}}{4}+2 x_{1}+\left(\frac{y_{1}-3}{2}\right)^{2}=0$
$\Rightarrow \quad \frac{x_{1}^{2}}{4}+2 x_{1}+\frac{y_{1}^{2}+9-6 y_{1}}{4}=0$
$\Rightarrow \quad x_{1}^{2}+y_{1}^{2}+8 x_{1}-6 y_{1}+9=0$
Hence, locus of a point is $x^{2}+y^{2}+8 x-6 y+9=0$
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