The length of the direct common tangent of the circles $x^{2}+y^{2}-4x-10y+28=0$ and $x^{2}+y^{2}+4x-6y+4=0$ is

The first circle x squared plus y squared minus 4x minus 10y plus 28 equals 0 has its center at (2, 5) and a radius of 1. The second circle x squared plus y squared plus 4x minus 6y plus 4 equals 0 has its center at (-2, 3) and a radius of 2. The distance between the centers is the square root of ((2 minus -2) squared plus (5 minus 3) squared), which equals the square root of 20. The length of the direct common tangent is found using the formula the square root of (distance squared minus (r1 + r2) squared), so we calculate the square root of (20 minus 9), which is the square root of 11.

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