Question

- Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their
centers is 4 cm. Find the length of the common chord.
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

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,  PQ ⊥ AB  

∴      (perpendicular from center to the chord, bisects the chord)

Let PR=xcm ⇒ 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