Two circles of radii `5\ c m\ a n d\ 3c m` intersect at two points and the distance between their centres is `4c mdot` Find the length of the common chord.
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, seg`PQ⊥chord AB`
`AR=RB=1/2AB`....perpendicular from center to the chord, bisects the chord
Let `PR=x cm`
so `RQ=(4−x)cm`
In`/_\ARP,AP^2 =AR^2+PR^2`
`AR^2=5^2−x^2`...(1)
In `/_\ARQ,AQ^2=AR^2+QR^2`
`AR^2=3^2−(4−x)^2`...(2)
`5^2−x^2=3^2−(4−x)^2`...from (1) & (2)
`25−x^2=9−(16−8x+x^2)`
`25−x^2=−7+8x−x^2`
`32=8x`
`x=4` substituting in equation(1) we get , `AR^2=25-16=9`
`AR=3cm`
`AB=2xxAR=2xx3=6`
`AB=6cm`So length of common chord is6 cm.

Explanation

To find the length of the common chord \( AB \) of two intersecting circles with given radii and distance between their centers, we can follow these steps:

### Step 1: Understand the Problem
We have two circles with radii \( AP = 5 \, \text{cm} \) and \( AQ = 3 \, \text{cm} \) respectively. The distance between their centers \( PQ \) is \( 4 \, \text{cm} \). We need to find the length of the common chord \( AB \).

### Step 2: Set Up the Geometry
Let \( P \) and \( Q \) be the centers of the two circles. The common chord \( AB \) is perpendicular to the line segment \( PQ \) at point \( R \), which is the midpoint of \( AB \). Thus, \( AR = RB = \frac{1}{2} AB \).

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