Two circles whose radii are equal to 4 and 8 | vedclass

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(A) Let the radii of the two circles be $r_1 = 4$ and $r_2 = 8$. Since the circles intersect at right angles,the distance between their centers $d = C

Vedclass · Accepted Answer

$\frac{16}{\sqrt{5}}$

Vedclass · Answer

(A) Let the radii of the two circles be $r_1 = 4$ and $r_2 = 8$. Since the circles intersect at right angles,the distance between their centers $d = C_1C_2$ is given by $d^2 = r_1^2 + r_2^2$.$d^2 = 4^2 + 8^2 = 16 + 64 = 80$.$d = \sqrt{80} = 4\sqrt{5}$.Let $P$ be one of the intersection points and $M$ be the projection of $P$ onto the line joining the centers $C_1C_2$. The length of the common chord is $PQ = 2PM$.In the right-angled triangle $	riangle PC_1C_2$,$PM$ is the altitude to the hypotenuse $C_1C_2$.The area of $	riangle PC_1C_2$ can be expressed as $\frac{1}{2} 	imes C_1P 	imes C_2P = \frac{1}{2} 	imes C_1C_2 	imes PM$.$4 	imes 8 = (4\sqrt{5}) 	imes PM$.$PM = \frac{32}{4\sqrt{5}} = \frac{8}{\sqrt{5}}$.Therefore,the length of the common chord $PQ = 2 	imes PM = 2 	imes \frac{8}{\sqrt{5}} = \frac{16}{\sqrt{5}}$.

Two circles whose radii are equal to $4$ and $8$ intersect at right angles. The length of their common chord is:

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