First slide

Analysis of two circles in circles

Question

The number of common tangents that can be drawn to the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0 and x^{2} + y^{2} + 2 x + 2 y + 1 = 0$ is

Moderate

Solution

The two circles are
$x^{2} + y^{2} - 4 x - 6 y - 3 = 0 and x^{2} + y^{2} + 2 x + 2 y + 1 = 0 .$
The coordinates of the centres and radii are:
Centres: $C_{1} (2, 3) C_{2} (- 1, - 1)$
Radii: $r_{1} = 4 r_{2} = 1$
Clearly $C_{1} C_{2} = 5 = r_{1} + r_{2}$
Therefore, there are 3 common tangents to the given circles

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