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

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Solution

The correct option is C 3
Number of common tangents depend on the position of the circle with respect to each other.
(i) If circles touch externally $\Rightarrow C_{1} C_{2} = r_{1} + r_{2}$ , 3 common tangents.
(ii) If circles touch internally $\Rightarrow C_{1} C_{2} = r_{2} - r_{1}$ , 1 common tangents
(iii) If circles do not touch each other, 4 common tangents.
Given equations of circles are
$x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ .....(i)
$x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ ....(ii)
Centre of circle (i) is $C_{1} (2, 3)$ and radius, $r_{1}$
$= \sqrt{4 + 9 + 12} = 5$
Centre of circle (ii) is $C_{2} (- 3, - 9)$ and radius, $r_{2}$
$= \sqrt{9 + 81 - 26}) = 8$
Now, $C_{1} C_{2} = \sqrt{(2 + 3)^{2} + (3 + 9)^{2}} \Rightarrow C_{1} C_{2} = \sqrt{5^{2} + 12^{2}} \Rightarrow C_{1} C_{2} = \sqrt{25 + 144} = 13 ∴ r_{1} + r_{2} = 5 + 8 = 13$
Also, $C_{1} C_{2} = r_{1} + r_{2}$
Thus, both circles touch each other externally. Hence, there are three common tangents.

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The number of common tangents to the circles x2+y2−4x−6y−12=0 and x2+y2+6x+18y+26=0 is

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