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KEAM 2010 Math Paper

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Question : 109 of 120

Marks: +1, -0

The line segment joining the points (4,7) and (-2,-1) is a diameter of a circle. If the circle intersects the

x

A

B

A B

is equal to

4
5
6
7
8

Solution:

\because (4,7)

and

(-2,-1)

are end points of circle.

\Rightarrow

Centre of circle

=\left(\;\frac{4-2}{2},\;\frac{7-1}{2}\right)=(1,3)

∴ Slope of

A C=\;\frac{-1-0}{-2-a}=-\;\frac{1}{2+a}

and slope of

A D=\;\frac{7-0}{4-a}=\;\frac{7}{4-a}

since,

\triangle CAD

is right angle as

C D

is diameter.

\Rightarrow \;\;

(slope of

A C

(slope of

A D)=-1

\Rightarrow \;\; \left(\;\frac{1}{2+a}\right)\left(\;\frac{7}{4-a}\right)=-1

\Rightarrow \;\; \frac{7}{-a^{2}+2a+8}=-1

\Rightarrow \;\; a^{2}-2a-15=0

\Rightarrow \;\; a=-3,5

Similarly,

\;\; b=-3,5

\therefore

Point

A=(-3,0)

and piont

B=(5,0)

\therefore \;\; A B=\sqrt{(-3-5)^{2}+(0-0)^{2}}

=\sqrt{(-8)^{2}}=8

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