Two circles whose radii are equal to 4 and 8 intersects at right angles. length of their

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10-1.Circle and System of Circles

normal

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

A

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

B

$8$

C

$4\sqrt 6$

D

$\frac{8 \sqrt 5}{5}$

Solution

$\mathrm{C}_{1} \mathrm{C}_{2}=\sqrt{16+64}$

$=4 \sqrt{5}$

$\mathrm{PM}=\mathrm{PC}_{1} \sin \theta$

$=\mathrm{PC}_{1} \times \frac{\mathrm{PC}_{2}}{\mathrm{C}_{1} \mathrm{C}_{2}}=4 \times \frac{8}{4 \sqrt{5}}=\frac{8}{\sqrt{5}}$

$\therefore \mathrm{PQ}=2(\mathrm{PM})=\frac{16}{\sqrt{5}}$

Standard 11

Mathematics

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