Two circles with centers O and O' of radii 6cm and 8 cm respectively intersect two points P and Q such that OP and O'P are tangents to the two circle. The length of the common chord PQ is

Explanation

Here, two circles are of radii OP=3cm and PO'=4cm.
These two circles intersect at P and Q.

Here, OP and PO' are two tangents drawn at point P.
`angleOPO'=90^(@)`
[tangent at any point of circle is perpendicular to radius through the point of conctact]
Join OO' and PN,
In right angled `DeltaOPO'`,
`(OO')^(2)=(OP)^(2)+(PO')^(2)` [by Pythagoras theroem]
i.e., `("Hypotenuse")^(2)=("Base")^(2)+("Perpendiu lar")^(2)`
`=(3)^(2)+(4)^(2)=25`
`rArrOO'=5cm`
Also, `PNbotOO'`
In right angled `DeltaOPN`
`(OP)^(2)=(ON)^(2)+(NP)^(2)` [by Pythagoras theorem]
`rArr(NP)^(2)=3^(2)-x^(2)=9-x^(2)......(i)`
and in right angled `DeltaPNO'`,
`(PO)^(2)=(PN)^(2)+(NO')^(2)` [by Pythagoras theorem]
`(4)^(2)=(PN)^(2)+(5-x)^(2)`
`(PN)^(2)=16-(5-x)^(2).....(ii)`
From Eqs. (i) and (ii),
`9-x^(2)=16-(5-x)^(2)`
`rArr7+x^(2)-(25+x^(2)-10x)=0`
`rArr10x=18`
`:.x=18`
Again, in right angled `DeltaOPN`,
`OP^(2)=(ON)^(2)+(NP)^(2)` [by Pythagoras theorem]
`rArr3^(2)=(1.8)^(2)+(NP)^(2)`
`rArr(NP)^(2)=9-3.24=5.76`
`:.(NP)=2.4`
:. Length of common chord, `PQ=2PN=2xx2.4=4.8cm`