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Home » 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.
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.
CBSE | Circles | Class 10 | Exercise 9.4 | Maths | NCERT Exemplar
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.
NCERT Exemplar Class 10 Maths Chapter 9 Ex. 9.4 Question 5

As per the inquiry,

Two circles with focuses O and O’ of radii 3 cm and 4 cm, individually meet at two focuses P and Q, to such an extent that OP and O’P are digressions to the two circles and PQ is a typical harmony.

To Find: Length of normal harmony PQ

∠OPO’ = 90° [Tangent at a point on the circle is opposite to the sweep through mark of contact]

So OPO is a right-calculated triangle at P

Utilizing Pythagoras in △ OPO’, we have

    \[\left( OO' \right)2=\text{ }\left( O'P \right)2+\text{ }\left( OP \right)2\]

    \[\left( OO' \right)2\text{ }=\text{ }\left( 4 \right)2\text{ }+\text{ }\left( 3 \right)2\]

    \[\left( OO' \right)2\text{ }=\text{ }25\]

OO’ = 5 cm

Let ON = x cm and NO’ = 5 – x cm

In right calculated triangle ONP

    \[\left( ON \right)2+\text{ }\left( PN \right)2=\text{ }\left( OP \right)2\]

    \[x2\text{ }+\text{ }\left( PN \right)2\text{ }=\text{ }\left( 3 \right)2\]

    \[\left( PN \right)2=\text{ }9\text{ }\text{ }x2\text{ }\left[ 1 \right]\]

In right calculated triangle O’NP

    \[\left( O'N \right)2\text{ }+\text{ }\left( PN \right)2=\text{ }\left( O'P \right)2\]

    \[\left( 5\text{ }\text{ }x \right)2\text{ }+\text{ }\left( PN \right)2\text{ }=\text{ }\left( 4 \right)2\]

    \[25\text{ }\text{ }10x\text{ }+\text{ }x2\text{ }+\text{ }\left( PN \right)2\text{ }=\text{ }16\]

    \[\left( PN \right)2\text{ }=\text{ }-\text{ }x2+\text{ }10x\text{ }\text{ }9\left[ 2 \right]\]

From [1] and [2]

    \[9\text{ }\text{ }x2\text{ }=\text{ }-\text{ }x2\text{ }+\text{ }10x\text{ }\text{ }9\]

10x = 18

x = 1.8

From (1) we have

    \[\left( PN \right)2\text{ }=\text{ }9\text{ }\text{ }\left( 1.8 \right)2\]

    \[=9\text{ }\text{ }3.24\text{ }=\text{ }5.76\]

PN = 2.4 cm

    \[PQ\text{ }=\text{ }2PN\text{ }=\text{ }2\left( 2.4 \right)\text{ }=\text{ }4.8\text{ }cm\]

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