The length of the common chord of the circles | vedclass

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(C) Given,$r_1 = 15$ units,$r_2 = 20$ units,and the distance between centers $C_1 C_2 = 25$ units.Note that $r_1^2 + r_2^2 = 15^2 + 20^2 = 225 + 400 =

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$24$

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(C) Given,$r_1 = 15$ units,$r_2 = 20$ units,and the distance between centers $C_1 C_2 = 25$ units.Note that $r_1^2 + r_2^2 = 15^2 + 20^2 = 225 + 400 = 625 = 25^2 = (C_1 C_2)^2$.Since the sum of the squares of the radii equals the square of the distance between the centers,the triangle $	riangle A C_1 C_2$ is a right-angled triangle with $\angle C_1 A C_2 = 90^\circ$.The common chord $AB$ is perpendicular to the line joining the centers $C_1 C_2$ at point $D$.In $	riangle A C_1 C_2$,the altitude $AD$ to the hypotenuse $C_1 C_2$ is given by $AD = \frac{r_1 	imes r_2}{C_1 C_2}$.$AD = \frac{15 	imes 20}{25} = \frac{300}{25} = 12$ units.The length of the common chord is $2 	imes AD = 2 	imes 12 = 24$ units.

The length of the common chord of the circles of radii $15$ and $20$,whose centers are $25$ units of distance apart,is

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