Question

Two circles with centers O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.

Answer 1

Given : Two circles with centers O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’P are tangents to the two circles and PQ is a common chord.

To Find : Length of common chord PQ

∠OPO' = 90° [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

So OPO is a right-angled triangle at P

Using Pythagoras in △ OPO', we have

(OO')²= (O'P)²+ (OP)²

(OO')² = (4)² + (3)²

(OO')² = 25

OO' = 5 cm

Let ON = x cm and NO' = 5 - x cm

In right angled triangle ONP

(ON)²+ (PN)²= (OP)²

x² + (PN)² = (3)²

(PN)²= 9 - x² [1]

In right angled triangle O'NP

(O'N)² + (PN)²= (O'P)²

(5 - x)² + (PN)² = (4)²

25 - 10x + x² + (PN)² = 16

(PN)² = -x²+ 10x - 9[2]

From [1] and [2]

9 - x² = -x² + 10x - 9

10x = 18

x = 1.8

From (1) we have

(PN)² = 9 - (1.8)²

=9 - 3.24 = 5.76

PN = 2.4 cm

PQ = 2PN = 2(2.4) = 4.8 cm

Answer 2

In ∆OO’P,

(O’O)² = OP² + O’P²

= 3² + 4²

= 9 + 16

(OO’)² = 25

∴ OO’ = 5 cm

Since the line joining the centres of two intersecting circles is perpendicular bisector of their common chord.

OR ⊥ PQ and PR = RQ

Let OR be x, then O’R = 5 – x again Let PR = RQ = y cm

In ∆ORP,

OP² = OR² + PR²

9 = x² + y²  ...(1)

In ∆O’RP,

O’P² = O’R² + PR²

16 = (5 – x)² + y²

16 = 25 + x² – 10x + y²

16 = x² + y² + 25 – 10x

16 = 9 + 25 – 10x  ...[From (1)]

16 = 34 – 10x

10x = 34 – 16 = 18

x = \(\frac{18}{10}\) = 1.8 cm

Substitute the value of x = 1.8 in (1)

9 = (1.8)² + y²

y² = 9 – 3.24

y² = 5.76

y = \(\sqrt{5.76}\) = 2.4 cm

Hence PQ = 2(2.4) = 4.8 cm

Length of the common chord PQ = 4.8 cm.