Question

Question 5
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

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

Here, OP and PO’ are two tangents drawn at point P.
$\angle$ OPO’ = ${90}^{\circ }$
[ tangents at any point of circle is perpendicular to radius through the point of contact]
Join OO’ and PN
In right angled $△$ OPO’
$(O{O}^{\prime }{)}^2=(OP{)}^2+(P{O}^{\prime }{)}^2$ [ by Pythagoras theorem]
i.e $(Hypotenuse{)}^2$ = $(Base{)}^2$ + $(Perpendicular{)}^2$
$=(3{)}^2+(4{)}^2=25$
$\Rightarrow$ OO’ = 5cm
Also. PN $\perp$ OO’
Let ON = x, then NO’ = 5 - x
In right angled $\angle$ OPN ,
$(OP{)}^2=(ON{)}^2+(NP{)}^2$ [ by Pythagoras theorem]
$\Rightarrow (NP{)}^2=3^2-x^2=9-x^2$
And in right angled $△$ PNO'
$(P{O}^{\prime }{)}^2=(PN{)}^2+(N{O}^{\prime }{)}^2$ [by Pythagoras theorem]
$\Rightarrow (4{)}^2=(PN{)}^2+(5-x{)}^2$
$\Rightarrow (PN{)}^2=16-(5-x{)}^2$
From Eqs. (i) and (ii)
$9-x^2=16-(5-x{)}^2\Rightarrow 7+x^2-(25+x^2-10x)=0\Rightarrow 10x=18\therefore x=1.8$
Again in right angled $△$ OPN
$O{P}^2=(ON{)}^2+(NP{)}^2$ [ by Pythagoras theorem]
$\Rightarrow 3^2=(1.8{)}^2+(NP{)}^2$
$\Rightarrow (NP{)}^2=9-3.24=5.76$
$\therefore$ (NP) = 2.4
$\therefore$ Length of common chord.
PQ = 2 PN = 2 $\times$ 2.4 = 4.8 cm.