ત્રણ વર્તુળ જેમની ત્રિજ્યા અનુક્રમે a, b, c, | vedclass

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Question

Length of direct common tangent for 

circle ${C_1}$  and ${C_2}$ is

$AB = \sqrt {{{\left( {a + b} \right)}^2} - {{\left( {a

Vedclass · Accepted Answer

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

Vedclass · Answer

Length of direct common tangent for 

circle ${C_1}$  and ${C_2}$ is

$AB = \sqrt {{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}} $

For ${C_3}$ and ${C_2}$

Length of direct common tangent for is

$BC = \sqrt {{{\left( {a + c} \right)}^2} - {{\left( {a - c} \right)}^2}} $

For ${C_1}$ and ${C_3}$ 

Length of direct common tangent for is

$AC = \sqrt {{{\left( {a + c} \right)}^2} - {{\left( {b - c} \right)}^2}} $

$AB + BC = AC$

$\sqrt {{{\left( {a + c} \right)}^2} - {{\left( {a - c} \right)}^2}}  + \sqrt {{{\left( {a + c} \right)}^2} - {{\left( {a - c} \right)}^2}} $

$ = \sqrt {{{\left( {a + c} \right)}^2} - {{\left( {b - c} \right)}^2}} $

$\sqrt {ab}  + \sqrt {ac}  = \sqrt {bc} $

$\frac{1}{{\sqrt c }} + \frac{1}{{\sqrt b }} = \frac{1}{{\sqrt a }}$

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