The length of the common chord of the circles | vedclass

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(C) Let the radii of the two circles be $r_1 = 15$ and $r_2 = 20$. The distance between their centers $C_1$ and $C_2$ is $d = 25$.Note that $r_1^2 + r

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$24$

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(C) Let the radii of the two circles be $r_1 = 15$ and $r_2 = 20$. The distance between their centers $C_1$ and $C_2$ is $d = 25$.Note that $r_1^2 + r_2^2 = 15^2 + 20^2 = 225 + 400 = 625 = 25^2 = d^2$.Since the sum of the squares of the radii equals the square of the distance between the centers,the triangle $	riangle AC_1C_2$ is a right-angled triangle with $\angle C_1AC_2 = 90^\circ$.The common chord $AB$ is perpendicular to the line joining the centers $C_1C_2$. Let the intersection point be $D$.In $	riangle AC_1C_2$,the altitude $AD$ to the hypotenuse $C_1C_2$ represents half the length of the common chord.Using the area of the triangle: $	ext{Area} = \frac{1}{2} 	imes r_1 	imes r_2 = \frac{1}{2} 	imes d 	imes AD$.$15 	imes 20 = 25 	imes AD$.$AD = \frac{300}{25} = 12$.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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