10-1.Circle and System of Circles
hard

एक वृत्त $C$, बिन्दु $(4,0)$ से होकर जाता है तथा वृत्त $x ^{2}+ y ^{2}+4 x -6 y =12$ को बिन्दु $(1,-1)$ पर बाह्य स्पर्श करता है, तो $C$ की त्रिज्या है

A

$2\sqrt 5 $

B

$4$

C

$5$

D

$\sqrt {57} $

(JEE MAIN-2019)

Solution

Tangent at $\left( {1, – 1} \right)$ is $x\left( 1 \right) + y\left( { – 1} \right) + 2\left( {x + 1} \right) – 3\left( {y – 1} \right) – 12 = 0$

$ = 3x – 4y = 7$

required circle is ${\left( {x + 1} \right)^2} + {\left( {y – 1} \right)^2} + \lambda \left( {3x – 4y – 7} \right) = 0$

It pass through $\left( {4,0} \right)$

$ \Rightarrow 9 + 1 + \lambda \left( {12 – 7} \right) = 0 \Rightarrow  =  – 2$

$ \Rightarrow $ required circleb is ${x^2} + {y^2} – 8x + 10y + 16 = 0$

$ \Rightarrow $ Radius $ = \sqrt {16 + 25 – 16}  = 5$

 

Standard 11
Mathematics

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