Two circles with centres 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.

#### Solution

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 ΔOPO’, 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}

x^{2} + (PN)^{2} = (3)^{2}

(PN)^{2}= 9 – x^{2} ......[1]

In right angled triangle O’NP

(O’N)^{2} + (PN)^{2}= (O’P)^{2}

(5 – x)^{2} + (PN)^{2} = (4)^{2}

25 – 10x + x^{2} + (PN)^{2} = 16

(PN)^{2} = – x^{2}+ 10x – 9 ......[2]

From [1] and [2]

9 – x^{2} = – x^{2} + 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