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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.

Answers (1)

Answer
4.8 cm
Solution 
              

Given : Radii of two circles are OP = 3 cm and {O}'  and intersection point of two circles are P and Q.  Here two tangents drawn at point P are OP and {O}'P
\therefore \: \angle P= 90^{\circ}
3^{2}-x^{2}= \left ( NP \right )^{2}
Also apply Pythagoras theorem in \bigtriangleup PN{O}'  we get
\left ( P{O}' \right )^{2}= \left ( PN \right )^{2}+\left (N {O}' \right )^{2}
\left (4 \right )^{2}= \left ( PN \right )^{2}+\left (5-x \right )^{2}
16= \left ( PN \right )^{2}+\left (5-x \right )^{2}
16- \left (5-x \right )^{2}= \left ( PN \right )^{2}\cdots \left ( ii \right )
Equate equation (i) and (ii) we get
9-x^{2}= 16-\left ( 5-x \right )^{2}
9-x^{2}-16+\left ( 25+x^{2}-10x \right )= 0
\left [ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right ]
-7-x^{2}+\left ( 25+x^{2}-10x \right )= 0
-7-x^{2}+25+x^{2}-10x= 0
18-10x= 0
18= 10x
\frac{18}{10}= x

x= 1\cdot 8
Put x = 1.8 in equation (i) we get  
9-\left ( 1\cdot 8 \right )^{2}= NP^{2}
9-3\cdot 24= NP^{2}
5\cdot 76= NP^{2}
NP= \sqrt{5\cdot 76}
NP= 2\cdot 4
P\mathbb{Q}= 2\times PN= 2\cdot 24= 4\cdot 8 cm  

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