Find the number of common tangents to the cir | vedclass

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(C) For the first circle $x^2 + y^2 + 2x + 8y - 23 = 0$:Center $C_1 = (-1, -4)$ and radius $r_1 = \sqrt{(-1)^2 + (-4)^2 - (-23)} = \sqrt{1 + 16 + 23}

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$2$

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(C) For the first circle $x^2 + y^2 + 2x + 8y - 23 = 0$:Center $C_1 = (-1, -4)$ and radius $r_1 = \sqrt{(-1)^2 + (-4)^2 - (-23)} = \sqrt{1 + 16 + 23} = \sqrt{40} = 2\sqrt{10} \approx 6.32$.For the second circle $x^2 + y^2 - 4x - 10y + 9 = 0$:Center $C_2 = (2, 5)$ and radius $r_2 = \sqrt{2^2 + 5^2 - 9} = \sqrt{4 + 25 - 9} = \sqrt{20} = 2\sqrt{5} \approx 4.47$.The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(2 - (-1))^2 + (5 - (-4))^2} = \sqrt{3^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10} \approx 9.49$.Now,check the relationship between $d$,$r_1 + r_2$,and $|r_1 - r_2|$:$r_1 + r_2 = 2\sqrt{10} + 2\sqrt{5} \approx 6.32 + 4.47 = 10.79$.$|r_1 - r_2| = 2\sqrt{10} - 2\sqrt{5} \approx 6.32 - 4.47 = 1.85$.Since $|r_1 - r_2| Therefore,there are $2$ common tangents.

Find the number of common tangents to the circles $x^2 + y^2 + 2x + 8y - 23 = 0$ and $x^2 + y^2 - 4x - 10y + 9 = 0$.

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