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A circle ${C_1}$ of radius $2$ touches both $x$ - axis and $y$ - axis. Another circle ${C_2}$ whose radius is greater than $2$ touches circle ${C_1}$ and both the axes. Then the radius of circle ${C_2}$ is
$6 - 4\sqrt 2 $
$6 + 4\sqrt 2 $
$6 - 4\sqrt 3 $
$6 + 4\sqrt 3 $
Solution
(b) First circle touches both axes and radius is $2$ unit.
Hence centre of circle is $(2, 2)$.
Let radius of other circle be a and this circle also touches both the axes.
Hence centre of circle is $(a, a).$
This circle touches first circle
Hence, $\sqrt {{{(a – 2)}^2} + {{(a – 2)}^2}} = a + 2$ squaring both the sides,${a^2} – 12a + 4 = 0$
$a = \frac{{12 \pm \sqrt {{{(12)}^2} – 4 \times 4 \times 1} }}{2}$
$ = \frac{{12 \pm \sqrt {128} }}{2} = 6 \pm 4\sqrt 2 $
But $a > 2$, Hence $a = 6 – 4\sqrt 2 $ is neglected
Hence, $a = 6 + 4\sqrt 2 $.
