Let a circle S = 0 touch both the circles x^2 | vedclass

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(D) Let the circle $S=0$ have center $(h, r)$ and radius $r$,where $r$ is the radius. Since it touches the $x$-axis,its center is $(h, r)$.Given circl

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$240$

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(D) Let the circle $S=0$ have center $(h, r)$ and radius $r$,where $r$ is the radius. Since it touches the $x$-axis,its center is $(h, r)$.Given circles are $C_1: x^2 + y^2 = 20^2$ (center $O_1(0,0)$,radius $R_1=20$) and $C_2: (x-5)^2 + (y-12)^2 = 5^2 + 12^2 - 120 = 169 - 120 = 49 = 7^2$ (center $O_2(5,12)$,radius $R_2=7$).Since $S$ touches $C_1$ externally,the distance between centers is $O_1C = R_1 + r$ $\Rightarrow \sqrt{h^2 + r^2} = 20 + r$ $\Rightarrow h^2 + r^2 = 400 + 40r + r^2$ $\Rightarrow h^2 = 400 + 40r$.Since $S$ touches $C_2$ externally,the distance between centers is $O_2C = R_2 + r$ $\Rightarrow \sqrt{(h-5)^2 + (r-12)^2} = 7 + r$ $\Rightarrow (h-5)^2 + (r-12)^2 = (7+r)^2$.Expanding: $h^2 - 10h + 25 + r^2 - 24r + 144 = 49 + 14r + r^2$.Substituting $h^2 = 400 + 40r$: $400 + 40r - 10h + 169 - 24r = 49 + 14r$ $\Rightarrow 520 - 10h = 14r$ $\Rightarrow 10h = 520 - 14r$ $\Rightarrow h = 52 - 1.4r$.Substituting $h$ back into $h^2 = 400 + 40r$: $(52 - 1.4r)^2 = 400 + 40r$.$2704 - 145.6r + 1.96r^2 = 400 + 40r \Rightarrow 1.96r^2 - 185.6r + 2304 = 0$.Solving this quadratic equation gives $r = 240$ or $r = 4.89$. Given the options,$r = 240$ is the correct radius.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

Let a circle $S = 0$ touch both the circles $x^2 + y^2 = 400$ and $x^2 + y^2 - 10x - 24y + 120 = 0$ externally and also touch the $x$-axis. The radius of the circle $S = 0$ is

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