If a circle C passing through (4, 0) touches circle x^2 + y^2 + 4x - 6y - 12 = 0 externally at a poi

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10-1.Circle and System of Circles

hard

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

A

$5$

B

$2\sqrt 5$

C

$4$

D

$\sqrt {57}$

(JEE MAIN-2013)

Solution

Let $A$ be the center og given circle and $B$ be the center of circle $C$.

${x^2} + {y^2} + 4x – 6y – 1 = 0$

$\therefore A = \left( { – 2,3} \right)$ and $B = \left( {g,f} \right)$

Now, from the figure, we have

$\frac{{ – 2 + g}}{2} = 1\,$ and $\frac{{3 + f}}{2} = – 1$

(By mid point formula)

$ \Rightarrow g = 4\,$ and $f = – 5$

Now, required radius

$ = OB = \sqrt {9 + 16} = \sqrt {25} = 5$

Standard 11

Mathematics

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