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10-1.Circle and System of Circles
hard
The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is
A
$4$
B
$1$
C
$2$
D
$3$
(JEE MAIN-2015)
Solution
$c_{1}(2,3);$
$r_{1}=\sqrt{4+9+12}=5$
And $c_{2}(-3,-9)$
$r_{2}=\sqrt{9+81-26}=8$
$\therefore \quad$ Distance $c_{1} c_{2}=\sqrt{25+144}=13$
$\therefore \quad c_{1} c_{2}=r_{1}+r_{2}$ touching externally.
$\Rightarrow \quad 3$ common tangents.
Standard 11
Mathematics
