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Maths / Circles / Concentric Circles Problems

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Concentric Circles Problems

Concentric Circles: The circle which have common centre point and different radius are called concentric circles.

Illustration: Two concentric circles with centre O have, A, B, C, D as the points of intersection with the line l as shown in the diagram. If AD = 12 cm and BC = 8 cm, find the lengths of AB, CD, AC and BD.

SOLUTION Since $large OM perp BC$

$large therefore ;;BM = CM = frac{1}{2}BC = 4cm$

Similarly, $large OM perp AD$

$large Rightarrow AM = DM = frac{1}{2}AD=6cm.$

Now, $large AB = AM - BM = (6-4)cm = 2cm$

Also, $large CD = DM-CM =(6-4)cm = 2cm$

$large therefore AC=AB+BC=(2+8)cm=10cm$

and, $large BD =BC+CD=(8+2)cm=10cm$

Illustration: Two concentric circles $C_1$ and $C_2$ with centre O have, A, B, C, D as the points of intersection with the line l as shown in the diagram. If AD = 8 cm and BC = 6 cm, and the distance of the chord from centre is 2 cm,find the length of radius of two circles.

SOLUTION Since $large OM perp BC$

$large therefore ;;BM = CM = frac{1}{2}BC = frac{6}{2}=3cm$

Similarly, $large OM perp AD$

$large Rightarrow AM = DM = frac{1}{2}AD=frac{8}{2}= 4cm.$

Now $large Delta$ AOM is a right angled at M

$large OA^2=OM^2+AM^2$ [ By pythagorous Theorem]

$large OA^2=2^2+4^2= 4 +16 = 20$

$large OA=sqrt {20} =2sqrt 5 cm$

Now $large Delta$ BOM is a right angled at M

$large OB^2=OM^2+BM^2$ [ By pythagorous Theorem]

$large OB^2=2^2+3^2= 4 +9 = 13$

$large OB=sqrt {13} cm$

Hence radius of circle $large C_1; is;sqrt {13} cm; and;C_2; is;2sqrt {5} cm;$

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Concentric Circles Problems

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