Test Index

Circles Part 1

Section: Mathematics

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Question : 80 of 100

Marks: +1, -0

Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point

(1,0)

to the distance from the point

(-1,0)

\;\frac{1}{3}

. Then the circumcentre of the triangle

A\ B\ C

is at the point:

$\left(\;\frac{5}{4}, 0\right)$
$\left(\;\frac{5}{2}, 0\right)$
$\left(\;\frac{5}{3}, 0\right)$
$(\;0, 0)$

Solution:

Given that

P(1,0), Q(-1,0)

and

\;\frac{A\ P}{A\ Q}=\;\frac{B\ P}{B\ Q}=\;\frac{C\ P}{C\ Q}=\;\frac{1}{3}

\Rightarrow 3 A P = A Q

Let

A = (x, y)

then

3 A P = A Q \Rightarrow 9 A P^2 = A Q^2

\Rightarrow 9(x-1)^2+9 y^2=(x+1)^2+y^2

\Rightarrow 9 x^2-18 x+9+9 y^2=x^2+2 x+1+y^2

\Rightarrow 8 x^2-20 x+8 y^2+8=0

\Rightarrow x^2+y^2-\;\frac{5}{3} x+1=0

......(1)

\therefore

A lies on the circle given by eq. (1). As

B

and

C

also follow the same condition, - they must lie on the same circle.

\therefore

Center of circumcircle of

\triangle A B C

=

Center of circle given by

(1) = \left(\;\frac{5}{4}, 0\right)

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