Q50 / 90 · JEE Main 2015 (04 Apr) · Mathematics

The number of common tangents to the circles $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ , is

A
$4$
B
$1$
C
$2$
Correct answer
D
$3$

Solution

Given

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

C_{1} (2, 3), r_{1} = \sqrt{2^{2} + 3^{2} + 12} = \sqrt{25} = 5

And

x^{2} + y^{2} + 6 x + 18 y + 26 = 0

C_{2} (- 3, - 9), r_{2} = \sqrt{3^{2} + 9^{2} - 26}

= \sqrt{90 - 26} = 8

Then,

C_{1} C_{2} = \sqrt{5^{2} + 1 2^{2}} = 13

C_{1} C_{2} = r_{1} + r_{2}

\Rightarrow

Externally touching circles.

\Rightarrow 3

common tangents.

Hence, the correct answer is option: 'D'.

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