Three circles of radii a, b, c (a

Question

(A) Let the radii of the three circles be $b, a, c$ respectively,where $a$ is the radius of the smallest circle placed between the two larger circles

Vedclass · Accepted Answer

$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}$

Vedclass · Answer

(A) Let the radii of the three circles be $b, a, c$ respectively,where $a$ is the radius of the smallest circle placed between the two larger circles of radii $b$ and $c$.The length of the direct common tangent between two circles of radii $r_1$ and $r_2$ touching each other externally is given by $L = \sqrt{(r_1+r_2)^2 - (r_1-r_2)^2} = 2\sqrt{r_1r_2}$.Let the points of contact of the circles with the $x$-axis be $A, B, C$ respectively.The distance $AB = 2\sqrt{ab}$ (between circles of radii $b$ and $a$).The distance $BC = 2\sqrt{ac}$ (between circles of radii $a$ and $c$).The distance $AC = 2\sqrt{bc}$ (between circles of radii $b$ and $c$).Since the smallest circle is between the other two,we have $AC = AB + BC$.$2\sqrt{bc} = 2\sqrt{ab} + 2\sqrt{ac}$.Dividing both sides by $2\sqrt{abc}$,we get:$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{b}}$.

Three circles of radii $a, b, c$ $(a < b < c)$ touch each other externally. If they have the $x$-axis as a common tangent,then:

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