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

Vedclass · Accepted Answer

$\frac{120}{13}$

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(B) Let the radii of the two circles be $r_1 = 5 \ cm$ and $r_2 = 12 \ cm$. Since the circles intersect at $90^{\circ}$,the distance between their centers $d$ is given by $d = \sqrt{r_1^2 + r_2^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \ cm$. Let the common chord be $AB$ and the centers of the circles be $O_1$ and $O_2$. The common chord $AB$ is perpendicular to the line joining the centers $O_1O_2$. Let $h$ be the distance from $O_1$ to the common chord $AB$. Then the distance from $O_2$ to $AB$ is $d - h = 13 - h$. In the right-angled triangles formed by the radii,the common chord,and the line of centers,we have: $r_1^2 - h^2 = r_2^2 - (13 - h)^2$ $25 - h^2 = 144 - (169 - 26h + h^2)$ $25 - h^2 = 144 - 169 + 26h - h^2$ $25 = -25 + 26h$ $50 = 26h \implies h = \frac{50}{26} = \frac{25}{13}$. The half-length of the common chord $x$ is given by $x = \sqrt{r_1^2 - h^2} = \sqrt{25 - (\frac{25}{13})^2} = \sqrt{25 - \frac{625}{169}} = \sqrt{\frac{4225 - 625}{169}} = \sqrt{\frac{3600}{169}} = \frac{60}{13}$. 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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