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(B) Let the radii of the two circles be $r_1 = 5$ and $r_2 = 12$. Since the circles intersect at $90^{\circ}$,the distance between their centers is $d

Vedclass · Accepted Answer

$\frac{120}{13}$

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(B) Let the radii of the two circles be $r_1 = 5$ and $r_2 = 12$. Since the circles intersect at $90^{\circ}$,the distance between their centers is $d = \sqrt{r_1^2 + r_2^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \ cm$.Let the common chord have length $2x$. The common chord is perpendicular to the line joining the centers.Let the common chord divide the line joining the centers into two segments of lengths $d_1$ and $d_2$,such that $d_1 + d_2 = 13$.In the two right-angled triangles formed by the radii,the common chord,and the segments of the line of centers,we have $x^2 + d_1^2 = 5^2$ and $x^2 + d_2^2 = 12^2$.Subtracting the equations: $d_2^2 - d_1^2 = 144 - 25 = 119$.Since $d_2 + d_1 = 13$,we have $d_2 - d_1 = \frac{119}{13}$.Adding the two equations for $d_1$ and $d_2$: $2d_2 = 13 + \frac{119}{13} = \frac{169 + 119}{13} = \frac{288}{13} \implies d_2 = \frac{144}{13}$.Then $d_1 = 13 - \frac{144}{13} = \frac{169 - 144}{13} = \frac{25}{13}$.Now,$x^2 = 5^2 - d_1^2 = 25 - (\frac{25}{13})^2 = 25(1 - \frac{25}{169}) = 25(\frac{144}{169})$.$x = 5 	imes \frac{12}{13} = \frac{60}{13}$.The length of the common chord is $2x = 2 	imes \frac{60}{13} = \frac{120}{13} \ cm$.

If the angle of intersection at a point where the two circles with radii $5 \ cm$ and $12 \ cm$ intersect is $90^{\circ}$,then the length (in $cm$) of their common chord is

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