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10-1.Circle and System of Circles
hard
If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^2 + y^2 + 4x -6y = 12$ externally at the point $(1, -1)$, then the radius of $C$ is
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
