The number of common tangent(s) to the circle | vedclass

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(C) Let $S_{1}: x^{2}+y^{2}+2x+8y-23=0$ and $S_{2}: x^{2}+y^{2}-4x-10y+19=0$.For $S_{1}$,the center $C_{1} = (-1, -4)$ and radius $r_{1} = \sqrt{(-1)^

Vedclass · Accepted Answer

$3$

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(C) Let $S_{1}: x^{2}+y^{2}+2x+8y-23=0$ and $S_{2}: x^{2}+y^{2}-4x-10y+19=0$.For $S_{1}$,the center $C_{1} = (-1, -4)$ and radius $r_{1} = \sqrt{(-1)^2 + (-4)^2 - (-23)} = \sqrt{1 + 16 + 23} = \sqrt{40} = 2\sqrt{10}$.For $S_{2}$,the center $C_{2} = (2, 5)$ and radius $r_{2} = \sqrt{2^2 + 5^2 - 19} = \sqrt{4 + 25 - 19} = \sqrt{10}$.The distance between the centers $C_{1}C_{2} = \sqrt{(2 - (-1))^2 + (5 - (-4))^2} = \sqrt{3^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$.Since $C_{1}C_{2} = r_{1} + r_{2}$ $(3\sqrt{10} = 2\sqrt{10} + \sqrt{10})$,the two circles touch each other externally.When two circles touch each other externally,there are exactly $3$ common tangents.

The number of common tangent$(s)$ to the circles $x^2 + y^2 + 2x + 8y - 23 = 0$ and $x^2 + y^2 - 4x - 10y + 19 = 0$ is:

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