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Standard X

Mathematics

Tangent Circles

Let C1 and ...

Question

Let $C_{1}$ and $C_{2}$ denote the centres of the circles $x^{2} + y^{2} = 4$ and $(x - 2)^{2} + y^{2} = 1$ respectively and let $P$ and $Q$ be their points of intersection. Then, the areas of $△ C_{1} P Q$ and $C_{2} P Q$ are in the ratio

3 : 1

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5 : 1

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7 : 1

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9 : 1

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Solution

The correct option is B $7 : 1$
$x^{2} + y^{2} = 4$ ...... (i)
$(x - 2)^{2} + y^{2} = 1$ ..... (ii)
$C_{1} = (0, 0), C_{2} = (2, 0)$

On solving Eqs. (i) and (ii), we get
$x = \frac{7}{4}, y = \pm \frac{\sqrt{15}}{4}$

$C_{1} N = \frac{7}{4}, P Q = \frac{2 \sqrt{15}}{4} = \frac{\sqrt{15}}{2}$

$\frac{A r e a (△ C_{1} P Q)}{A r e a (△ C_{2} P Q)} = \frac{\frac{1}{2} \cdot P Q \cdot C_{1} N}{\frac{1}{2} \cdot P Q \cdot C_{2} N}$

$= \frac{C_{1} N}{C_{2} N} = \frac{\frac{7}{4}}{2 - \frac{7}{4}} = \frac{\frac{7}{4}}{\frac{1}{4}} = 7 : 1$

Hence, the required option is (c).

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Let C1 and C2 denote the centres of the circles x2+y2=4 and (x−2)2+y2=1 respectively and let P and Q be their points of intersection. Then, the areas of △C1PQ and C2PQ are in the ratio

The correct option is B 7:1x2+y2=4 ...... (i)(x−2)2+y2=1 ..... (ii)C1=(0,0),C2=(2,0)On solving Eqs. (i) and (ii), we getx=74,y=±√154C1N=74,PQ=2√154=√152Area(△C1PQ)Area(△C2PQ)=12⋅PQ⋅C1N12⋅PQ⋅C2N=C1NC2N=742−74=7414=7:1Hence, the required option is (c).

Let $C_{1}$ and $C_{2}$ denote the centres of the circles $x^{2} + y^{2} = 4$ and $(x - 2)^{2} + y^{2} = 1$ respectively and let $P$ and $Q$ be their points of intersection. Then, the areas of $△ C_{1} P Q$ and $C_{2} P Q$ are in the ratio