Question

The equation of a circle of equal radius, touching both the circles $x^2+y^2=a^2$ and ${(x-2a)}^2+y^2=a^2$ is given by

• A. $x^2+y^2-2ax-2\sqrt{3}ay+3a^2=0$
• B. $x^2+y^2-2ax+2\sqrt{3}ay+3a^2=0$
• C. $x^2+y^2+2ax-2\sqrt{3}ay+3a^2=0$
• D. None of these

Answer 1

Answer: (A), (B)

$x^2+y^2-2ax-2\sqrt{3}ay+3a^2=0$
$x^2+y^2-2ax+2\sqrt{3}ay+3a^2=0$

Given circles are $x^2+y^2=a^2$ ...(1)

and, ${(x-2a)}^2+y^2=a^2$ ...(2)

Let $A$ and $B$ be the centres and $r_1$ and $r_2$ the radii of the two circles (1) and (2), respectively, then

$A\equiv (0,0),B\equiv (2a,0),r_1=a,r_2=a$

Let the equation of equal circle touching circles (1) and (2) be

${(x-\alpha )}^2+{(y-\beta )}^2=a^2$ ...(3)

Its centre $C$ is $(\alpha ,\beta )$ and radius $r_3=a.$

Since (3) touches (1),

$\therefore AC=r_1+r_3=2a$

$[$ Here $AC\ne |r_1-r_3|$ as $r_1-r_3=a-a=0]$

$\Rightarrow {AC}^2=4a^2$ $\Rightarrow {\alpha }^2+{\beta }^2={4a}^2$ ...(4)

Again, since circle (3) touches (2),

$\therefore BC=r_2+r_3{BC}^2={(r_2+r_3)}^2$

$\therefore {(2a-\alpha )}^2+{\beta }^2={(a+a)}^2{\alpha }^2+{\beta }^2-4a\alpha =0$

$\Rightarrow {4a}^2-4a\alpha =0$ [from (4)]

$\therefore \alpha =a$ and from (4),

$\beta =\pm \sqrt{3}a.$

Thus, required circles are ${(x-a)}^2+{(y\mp \sqrt{3}a)}^2=a^2.$

$\Rightarrow x^2+y^2-2ax\mp 2\sqrt{3}ay+3a^2=0.$