Let a circle $ C_{1} $ be obtained on rolling the circle $ x^{2}+ $ $ y^{2}-4 x-6 y+11=0 $ upwards 4 units on the tangent $ T $ to it at the point $ (3,2) $. Let $ C_{2} $ be the image of $ C_{1} $ in $ T $. Let $ A $ and $ B $ be the centers of circles $ C_{1} $ and $ C_{2} $ respectively, and $ M $ and $ N $ be respectively the feet of perpendiculars drawn from $ A $ and $ B $ on the $ x $-axis. Then the area of the trapezium AMNB is:

JEE MATHEMATICS MCQ: CIRCLES

1 Easy

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Analyzing +4 | -1

Let a circle

C_{1}

be obtained on rolling the circle

x^{2}+

y^{2}-4 x-6 y+11=0

upwards 4 units on the tangent

T

to it at the point

(3,2)

. Let

C_{2}

be the image of

C_{1}

T

. Let

A

and

B

be the centers of circles

C_{1}

and

C_{2}

respectively, and

M

and

N

be respectively the feet of perpendiculars drawn from

A

and

B

on the

x

-axis. Then the area of the trapezium AMNB is:

View answer and explanation

Correct option: C

4(1+\sqrt{2})

Option C (

4(1+\sqrt{2})

) is correct because the area of the trapezium is calculated by finding the coordinates of the centers of the circles and using the formula for the area of a trapezium.

JEE MATHEMATICS MCQ: CIRCLES

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