Question

A chord $AB$ is drawn from the point $A(0,3)$ on the circle $x^2+4x+(y-3)2=0$, and is extended to $M$ such that $AM=2AB$. The locus of $M$ is

• A. $x^2+y^2-8x-6y+9=0$
• B. $x^2+y^2+8x+6y+9=0$
• C. $x^2+y^2+8x-6y+9=0$
• D. $x^2+y^2-8x+6y+9=0$

Answer 1

Answer: (C)

Given, $AM=2AB$

$\Rightarrow B$ is mid-point of $AM$.
$\therefore$ 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 $x^2+4x+(y-3{)}^2=0$
$\therefore {\left(\frac{x_1}{2}\right)}^2+4\left(\frac{x_1}{2}\right)+{\left(\frac{y_1+3}{2}-3\right)}^2=0$
$\Rightarrow \frac{x_1^2}{4}+2x_1+{\left(\frac{y_1-3}{2}\right)}^2=0$
$\Rightarrow \frac{x_1^2}{4}+2x_1+\frac{y_1^2+9-6y_1}{4}=0$
$\Rightarrow x_1^2+y_1^2+8x_1-6y_1+9=0$
Hence, locus of a point is
$x^2+y^2+8x-6y+9=0$