Two intersecting circles have a common chord of length 16 ft, and their centers lie on opposite sides of the chord. The radii of the circles are 10 ft and 17 ft respectively. Express the distance between the centers of the circles in feet.

Logic Jun 22, 2019

#1**+1 **

Let C_{1} be the center of the circle with radius 10 ft, C_{2} be the center of the circle with radius 17 ft, A and B be the endpoints of the common chord.

\(\angle \text{C}_1\text{AB} = \arccos\left(\dfrac{\dfrac{16}{2}}{10}\right) = \arccos\left(\dfrac{4}{5}\right)\\ \angle \text{C}_2\text{AB} = \arccos\left(\dfrac{\dfrac{16}{2}}{17}\right) = \arccos\left(\dfrac{8}{17}\right)\\ \text{Distance} = \sqrt{10^2 + 17^2 - 2(10)(17)\cos\left(\arccos\left(\dfrac{4}{5}\right)+\arccos\left(\dfrac{8}{17}\right)\right)} = 21\text{ ft.}\)

MaxWong Jun 22, 2019

#2**+1 **

We just use the distance formula: For lines, YA and AB, draw a diagram and label all the points!

Thus, we have \(\sqrt{{10^2}-{8^2}}=\sqrt{100-64}=\sqrt{36}=6\) feet.

Similarly, we do it for XA and XB, and since this is \(17\) feet, we get \(\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15\) feet.

Thus, the answer is \(6+15=\boxed{21}\) feet.

tertre Jun 22, 2019