Question

A circle touches both the circles $x^2+y^2=25$ and $(x-2{)}^2+y^2=1.$ Then locus of its centre is

• A. an ellipse with focus $(2,0)$ and $(0,10)$
• B. an ellipse with length of major axis $6$ unit
• C. an ellipse with eccentricity $\frac{1}{4}$
• D. an ellipse with auxiliary circle $x^2+y^2-2x-8=0$

Answer 1

Answer: (B), (D)

an ellipse with auxiliary circle $x^2+y^2-2x-8=0$

$C_1:x^2+y^2=25$
$C_2:(x-2{)}^2+y^2=1$
Clearly $C_2$ lies inside $C_1.$


Let coordinates of centre of variable circle be $(h,k)$ and $r$ be the radius.
Then $(h-2{)}^2+k^2=(1+r{)}^2\dots \left(1\right)$
and $h^2+k^2=(5-r{)}^2\dots (2)$

From $\left(1\right)-\left(2\right),$ we get
$(h-2{)}^2-h^2=(1+r{)}^2-(5-r{)}^2$
$\Rightarrow 4-4h=1+2r-25+10r$
$\Rightarrow 28=12r+4h$
$\Rightarrow r=\frac{7-h}{3}$

Putting in equation $\left(2\right),$ we get
$h^2+k^2={(5-\frac{7-h}{3})}^2$
$\Rightarrow 8(h-1{)}^2+9k^2=72$
Hence, required locus is
$\frac{(x-1{)}^2}{9}+\frac{y^2}{8}=1$
which is an ellipse centred at $(1,0)$ with $a=3$ and $b=2\sqrt{2}$
Eccentricity, $e=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{3}$

Coordinates of foci are $(1\pm ae,0)$ i.e., $(2,0)$ and $(0,0).$

Equation of auxiliary circle of the ellipse is
$(x-1{)}^2+y^2=9$
$\Rightarrow x^2+y^2-2x-8=0$

Length of major axis $=2\times 3=6$