- Home
- Standard 11
- Mathematics
$a , b , c ( a < b < c )$ त्रिज्याओं वाले तीन वृत्त परस्पर बाह्य स्पर्श करते हैं। यदि $x$ -अक्ष उनकी एक उभयनिष्ठ स्पर्श रेखा है, तो :
$\frac{1}{{\sqrt a }} = \frac{1}{{\sqrt b }} + \frac{1}{{\sqrt c }}$
$\frac{1}{{\sqrt b }} = \frac{1}{{\sqrt a }} + \frac{1}{{\sqrt c }}$
$a, b, c$ स. श्रे. में है
$\sqrt a ,\sqrt b ,\sqrt c $ स. श्रे. में है
Solution
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} $
$\frac{1}{{\sqrt c }} + \frac{1}{{\sqrt b }} = \frac{1}{{\sqrt a }}$
