Question

# The locus of the centers of the circles which touch the two circle ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$and${{x}^{2}}+{{y}^{2}}=4ax$ externally is.A. $12{{x}^{2}}-4{{y}^{2}}-24ax+9{{a}^{2}}=0$B. $12{{x}^{2}}+4{{y}^{2}}-24ax+9{{a}^{2}}=0$C. $12{{x}^{2}}-4{{y}^{2}}+24ax+9{{a}^{2}}=0$D. $12{{x}^{2}}+4{{y}^{2}}+24ax+9{{a}^{2}}=0$

Hint: Take the equations and write their centers and radius. Then consider the center of a circle in which you want to find the locus of the center. Then equate it and solve it. You will get the answer.

A locus is the set of all points (usually forming a curve or surface) satisfying some condition. For example, the locus of points in the plane equidistant from a given point is a circle, and the set of points in three-space equidistant from a given point is a sphere.

A locus of points usually results in a curve or surface. For instance, in our hiking example, the locus of points 5 miles from our starting point resulted in a curve that's a circle.

Now, how do we usually represent curves algebraically? If you're thinking we use an equation, you're exactly right.
Let $A$ and $B$ are the given two circles with radii ${{R}_{1}}$ and ${{R}_{2}}$â€‹ respectively and their centers are $F$ and $G$respectively.
Let $C$ and $D$ are variable circles so that each circle meets the given circles $A$ and $B$ externally.

Let ${{r}_{1}}$â€‹ be the radius of the circle $C$ and $P$ be its center. Let ${{r}_{2}}$â€‹ be the radius of the circle $D$ and $Q$ be its center.
$PF={{R}_{1}}+{{r}_{1}},PG={{R}_{2}}+{{r}_{1}}$
Now subtracting above both we get,
$PF-PG={{R}_{1}}-{{R}_{2}}$
Also $QF={{R}_{1}}+{{r}_{2}},QG={{R}_{2}}+{{r}_{2}}$
Again subtracting we get,
$QF-QG={{R}_{1}}-{{R}_{2}}$

Hence locus of centers of touching circles is a set of points so that the difference between the distances from two given fixed points ( here $F$and $G$ ) to the point in locus is constant.
This locus is hyperbola and the fixed points$F$and $G$ (Centers of given circles ) are foci of hyperbola.

Now we have given ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$,
So let the center be${{C}_{1}}$,
${{C}_{1}}(0,0)$because we can see the coefficient of$x$and$y$is zero.
And radius ${{r}_{1}}=\sqrt{a}$.

Now again one equation is given that${{x}^{2}}+{{y}^{2}}=4ax$,
So let the center be${{C}_{2}}$,
${{C}_{2}}(2a,0)$because we can see a coefficient of$y$is zero.
And radius ${{r}_{2}}=\sqrt{{{g}^{2}}+{{f}^{2}}+{{c}^{2}}}=\sqrt{0+0+{{(2a)}^{2}}}=2a$.

Let the radius of the circle of which we want to find a locus be$r$.
And the center is$C(h,k)$.
So now using distance formula we get,
\begin{align} & r+a=\sqrt{{{h}^{2}}+{{(k-a)}^{2}}} \\ & r+2a=\sqrt{{{h}^{2}}+{{(k-2a)}^{2}}} \\ \end{align}

Let${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ be the variable circle.
Since it touches the given circle easily,
$\sqrt{{{(-g-0)}^{2}}+{{(-f-0)}^{2}}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}+a$ ...(1)
and, $\sqrt{{{(-g-2a)}^{2}}+{{(-f-0)}^{2}}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}+2a$ ...(2)
Subtracting (1) from (2), we get
$\sqrt{{{(g+2a)}^{2}}+{{f}^{2}}}=\sqrt{{{g}^{2}}+{{f}^{2}}}+a$
Squaring both sides, we get
${{(g+2a)}^{2}}+{{f}^{2}}={{(\sqrt{{{g}^{2}}+{{f}^{2}}}+a)}^{2}}$

\begin{align} & {{g}^{2}}+4ga+4{{a}^{2}}+{{f}^{2}}={{g}^{2}}+{{f}^{2}}+2a\sqrt{{{g}^{2}}+{{f}^{2}}}+{{a}^{2}} \\ & 4ag+3{{a}^{2}}=2a\sqrt{{{g}^{2}}+{{f}^{2}}} \\ \end{align}
Now divide the whole equation by $a$we get,
$4g+3a=2\sqrt{{{g}^{2}}+{{f}^{2}}}$
Now square both sides we get,
${{\left( 4g+3a \right)}^{2}}={{\left( 2\sqrt{{{g}^{2}}+{{f}^{2}}} \right)}^{2}}$
${{\left( (-4)(-g)+3a \right)}^{2}}=4\left( {{g}^{2}}+{{f}^{2}} \right)$
So the locus of the centre$(-g,-f)$is${{(-4x+3a)}^{2}}=4({{x}^{2}}+{{y}^{2}})$,
So simplifying above we get,

\begin{align} & 16{{x}^{2}}+9{{a}^{2}}-24ax=4{{x}^{2}}+4{{y}^{2}} \\ & 12{{x}^{2}}-4{{y}^{2}}-24ax+9{{a}^{2}}=0 \\ \end{align}
So we get locus of the centre as$12{{x}^{2}}-4{{y}^{2}}-24ax+9{{a}^{2}}=0$.
So the correct answer is an option(A).

Note: Read the question carefully. You should know the concept of locus of point and locus of the center. Donâ€™t jumble yourself in transferring the equations. Most of the students make mistakes in minus signs, So avoid the mistake.