The number of common tangents to the two circ | vedclass

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(B) The equations of the circles are $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$. For the first circle,the center $C_1 = (4, -1)$ and the radius $r_1

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The equations of the circles are $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$. For the first circle,the center $C_1 = (4, -1)$ and the radius $r_1 = \sqrt{4^2 + (-1)^2 - 0} = \sqrt{17}$. For the second circle,the center $C_2 = (1, 8)$ and the radius $r_2 = \sqrt{1^2 + 8^2 - 25} = \sqrt{1 + 64 - 25} = \sqrt{40}$. The distance between the centers is $C_1C_2 = \sqrt{(1-4)^2 + (8 - (-1))^2} = \sqrt{(-3)^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90}$. We compare $C_1C_2$ with the sum and difference of the radii: $r_1 + r_2 = \sqrt{17} + \sqrt{40} \approx 4.12 + 6.32 = 10.44$. $|r_1 - r_2| = |\sqrt{17} - \sqrt{40}| \approx |4.12 - 6.32| = 2.20$. Since $|r_1 - r_2| Therefore,the number of common tangents is $2$.

The number of common tangents to the two circles $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$ is:

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