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(D) Given,the slope of the diameter $AC$ is $m = \frac{3}{4}$.Since $	an 	heta = \frac{3}{4}$,we have $\cos 	heta = \frac{4}{5}$ and $\sin 	heta =

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$\frac{23}{13}$

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(D) Given,the slope of the diameter $AC$ is $m = \frac{3}{4}$.Since $	an 	heta = \frac{3}{4}$,we have $\cos 	heta = \frac{4}{5}$ and $\sin 	heta = \frac{3}{5}$.The centre of the circle is $(h, k) = (3, 2)$ and the radius is $r = 25$.The coordinates of the endpoints of the diameter are given by $(h \pm r \cos 	heta, k \pm r \sin 	heta)$.For $C = (x_2, y_2)$,using the positive sign:$x_2 = 3 + 25 	imes \frac{4}{5} = 3 + 20 = 23$$y_2 = 2 + 25 	imes \frac{3}{5} = 2 + 15 = 17$For $A = (x_1, y_1)$,using the negative sign:$x_1 = 3 - 25 	imes \frac{4}{5} = 3 - 20 = -17$$y_1 = 2 - 25 	imes \frac{3}{5} = 2 - 15 = -13$Therefore,$\frac{x_1 x_2}{y_1 y_2} = \frac{(-17) 	imes 23}{(-13) 	imes 17} = \frac{-17 	imes 23}{-13 	imes 17} = \frac{23}{13}$.

Let the slope of a diameter $AC$ of a circle of radius $25$ units be $\frac{3}{4}$. If $(3, 2)$ is the centre of the circle,$A = (x_1, y_1)$ and $C = (x_2, y_2)$,then $\frac{x_1 x_2}{y_1 y_2} = $

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