Analysis of two circles in circles

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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

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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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Analysis of two circles in circles

Remember concepts with our Masterclasses.

The number of common tangents that can be drawn to the circle x2+y2−4x−6y−3=0 and x2+y2+2x+2y+1=0 is

The two circles are x2+y2−4x−6y−3=0 and x2+y2+2x+2y+1=0. The coordinates of the centres and radii are:Centres: C1(2,3) C2(−1,−1)Radii: r1=4 r2=1Clearly C1C2=5=r1+r2Therefore, there are 3 common tangents to the given circles

Ready to Test Your Skills?

free study material

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