Question

From the point \[A(0,3)\] on the circle${x^2} + 4x + (y - 3)^2$, a chord \[AB\] is drawn and extended to a point \[M\], such that$AM = 2AB$. An equation of the locus \[M\] is
A) ${x^2} + 6x + {(y - 2)^2} = 0$
B) ${x^2} + 8x + {(y - 3)^2} = 0$
C) ${x^2} + {y^2} + 8x - 6y + 9 = 0$
D) ${x^2} + {y^2} + 6x - 4y + 4 = 0$

Answer 1

Hint: Use formula of equation of circle ${(x - a)^2} + {(y - b)^2} = {r^2}$
Where \[(x,y)\]the point on the circle is, $(a,b)$ is the coordinate of the center of the circle and $r$ is the radius of the circle.

Complete step-by-step answer:

Equation of circle is ${x^2} + 4x + {(y - 3)^2} = 0$
$AM = 2AB$
\[B\] is the midpoint of \[AM\]
Therefore,
$ \Rightarrow B = \left( {\dfrac{h}{2} + \dfrac{{k + 3}}{2}} \right)$ Lies on the circle
Now,
Equation of circle is ${x^2} + 4x + {(y - 3)^2} = 0$
Let, $x = \dfrac{h}{2},y = \dfrac{{k + 3}}{2}$
Therefore,
$\dfrac{{{h^2}}}{4} + 2h + {\left( {\dfrac{{k + 3}}{2} - 3} \right)^2} = 0$ (Putting the values of \[x\] and \[y\] in equation of the circle)
By solving above equation,
$ \Rightarrow \dfrac{{{h^2}}}{4} + 2h + \dfrac{{{k^2} - 6k + 9}}{4} = 0$
Therefore,
${k^2} + {h^2} + 8h - 6k + 9 = 0$
$\therefore $ Locus of \[M\] is ${x^2} + {y^2} + 8x - 6y + 9 = 0$
So, the correct answer is “Option C”.

Note: In this type of Coordinate Geometry Use carefully the coordinates of various points.Remember the Various equations of circle like${x^2} + {y^2} = {r^2}$ (when center is at the origin),
${(x - a)^2} + {(y - b)^2} = {r^2}$When center is any point on the Cartesian plane.