The point of intersection of the direct commo | vedclass

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Question

(D) The direct common tangents to two circles meet at the external center of similitude,which divides the line segment joining the centers externally

Vedclass · Accepted Answer

$(22,-4)$

Vedclass · Answer

(D) The direct common tangents to two circles meet at the external center of similitude,which divides the line segment joining the centers externally in the ratio of their radii $r_1 : r_2$.For the given circles $(x+11)^2+(y-2)^2=15^2$ and $(x-11)^2+(y+2)^2=5^2$,the centers are $C_1 = (-11, 2)$ and $C_2 = (11, -2)$,and the radii are $r_1 = 15$ and $r_2 = 5$.The ratio of the radii is $r_1 : r_2 = 15 : 5 = 3 : 1$.The external center of similitude $(x, y)$ is given by the section formula for external division:$x = \frac{r_1 x_2 - r_2 x_1}{r_1 - r_2} = \frac{15(11) - 5(-11)}{15 - 5} = \frac{165 + 55}{10} = \frac{220}{10} = 22$$y = \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2} = \frac{15(-2) - 5(2)}{15 - 5} = \frac{-30 - 10}{10} = \frac{-40}{10} = -4$Thus,the point of intersection is $(22, -4)$.

The point of intersection of the direct common tangents drawn to the circles $(x+11)^2+(y-2)^2=225$ and $(x-11)^2+(y+2)^2=25$ is

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