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

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

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

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(A) Let the radii of the two circles be $r_1 = 8$ and $r_2 = 4$. Since the circles intersect at right angles,the distance $d$ between their centers is given by $d = \sqrt{r_1^2 + r_2^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$.The common chord is perpendicular to the line joining the centers. Let $p$ be the length of the perpendicular from the center of one circle to the common chord. In the right-angled triangle formed by the radii and the distance between centers,the area can be expressed as $\frac{1}{2} 	imes r_1 	imes r_2 = \frac{1}{2} 	imes d 	imes p$.Substituting the values: $\frac{1}{2} 	imes 8 	imes 4 = \frac{1}{2} 	imes 4\sqrt{5} 	imes p$.$32 = 4\sqrt{5} 	imes p \Rightarrow p = \frac{32}{4\sqrt{5}} = \frac{8}{\sqrt{5}}$.The length of the common chord is $2p = 2 	imes \frac{8}{\sqrt{5}} = \frac{16}{\sqrt{5}}$.

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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