Question

The length of the common chord AB of two intersecting circles is 18 cm. If the diameters of the circles are 82 cm and 30 cm, calculate the distance between the centres.

Answer 1

Given:

• Length of the common chord AB = 18 cm
• Diameter of the first circle = 82 cm, so radius r₁ = `82/2` = 41 cm
• Diameter of the second circle = 30 cm, so radius r₂ = `30/2` = 15 cm
• The distance between the centers of the two circles P and Q is denoted as PQ.

1. Let the centers of the two circles be P and Q.

2. Let M be the midpoint of the chord AB.

Since AB is a common chord of length 18 cm, each segment:

AM = MB

= `18/2`

= 9 cm

3. The common chord is perpendicular to the line segment joining the centers PQ, so PQ ⊥ AB and M lies on PQ.

4. In ΔPAM: 

`PM = sqrt(r_1^2 - AM^2)`

= `sqrt(41^2 - 9^2)`

= `sqrt(1681 - 81)`

= `sqrt(1600)`

= 40 cm

5. In ΔQAM:

`QM = sqrt(r_2^2 - AM^2)`

= `sqrt(15^2 - 9^2)`

= `sqrt(225 - 81)`

= `sqrt(144)`

= 12 cm

6. Therefore, the distance between the centers is:

PQ = PM + QM

= 40 + 12

= 52 cm

The distance between the centers of the two circles is 52 cm.