Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 3

We know that,

If two circles touch internally, then distance between their centres is equal to the difference of their radii. So,

$AB\text{ }=\text{ }\left( 5\text{ }-\text{ }3 \right)\text{ }cm\text{ }=\text{ }2\text{ }cm$

Also, the common chord

$PQ$

is the perpendicular bisector of

$AB$

Thus,

$AC\text{ }=\text{ }CB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }AB\text{ }=\text{ }1\text{ }cm$

In right

$\vartriangle ACP$

we have

$A{{P}^{2}}~=\text{ }A{{C}^{2}}~+\text{ }C{{P}^{2}}~$

[Pythagoras Theorem]

${{5}^{2}}~=\text{ }{{1}^{2}}~+\text{ }C{{P}^{2}}$

Or,

$C{{P}^{2}}~=\text{ }25\text{ }-\text{ }1\text{ }=\text{ }24$

$CP\text{ }=\text{ }\surd 24\text{ }cm\text{ }=\text{ }2\surd 6\text{ }cm$

Now,

$PQ\text{ }=\text{ }2\text{ }CP$

$=\text{ }2\text{ }x~2\surd 6\text{ }cm$

So,

$=\text{ }4\surd 6\text{ }cm$

Therefore, the length of

$PQ\text{ }is\text{ }4\surd 6\text{ }cm$

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