Question

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Two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm. Find the length of the common chord.​

Answer 1

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Let the common chord be AB and P and Q be the centers of the two circles.

∴AP=5cm and AQ=3cm.

PQ=4cm ....given

Now, segPQ⊥chord AB

$∴AR=RB= \frac{1}{2} AB$

....perpendicular from center to the chord, bisects the chord

Let PR=xcm, so RQ=(4−x)cm

In △ARP,

AP² = AR² + PR²

AR² = 5² − x² ...(1)

In △ARQ,

AQ² = AR² + QR²

AR² =3² − ( 4 − x )² ...(2)

∴5² − x² = 3² − (4 − x)²....from (1) & (2)

25 − x² = 9 − (16 − 8x + x² )

25 − x² = − 7 + 8x − x²

32=8x

∴x=4

Substitute in eq(1) we get,

AR² =25−16=9

∴AR = 3cm.

∴AB = 2 × AR = 2×3

∴AB = 6cm.

So, length of common chord AB is 6cm.

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