a , b , c ( a

Question

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

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

$a , b , c ( a < b < c )$ त्रिज्याओं वाले तीन वृत्त परस्पर बाह्य स्पर्श करते हैं। यदि $x$ -अक्ष उनकी एक उभयनिष्ठ स्पर्श रेखा है, तो :

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