lf the circles $\mathrm{x}^{2}+\mathrm{y}^{2}-10\mathrm{x}+2\mathrm{y}+10=0$ and $\mathrm{x}^{2}+\mathrm{y}^{2}-4\mathrm{x}-6\mathrm{y}-12=0$ touch each other then the slope of the common tangent at the point of contact of the circles is

To find the slope of the common tangent, we first determine the centers and radii of the two circles: C1 is (5, -1) with radius 4, and C2 is (2, 3) with radius 5. The distance between centers C1 and C2 is exactly 5, which equals the difference of their radii (5 minus 4), meaning the circles touch internally. The common normal at the point of contact is the line joining C1 and C2, which has a slope of negative 4 thirds. Because the tangent is perpendicular to this normal, its slope is the negative reciprocal, which is 3 fourths.

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