JEE Main · 2009 · 12 Apr 2009 · Offline · Circle

Three distinct points

A, B

and C are given in the

2 -

dimensional coordinate 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

) is equal to

\frac{1}{3}

. Then the circumcentre of the triangle

A B C

is at the point

(A) $(0, 0)$
(B) $(\frac{5}{4}, 0)$ ✓ Correct
(C) $(\frac{5}{2}, 0)$
(D) $(\frac{5}{3}, 0)$

Answer: (B)

Solution

Writing a general point as

(x, y)

, the ratio condition

\frac{A P}{A Q} = \frac{1}{3}

with

P = (1, 0)

and

Q = (- 1, 0)

becomes

9 [(x - 1)^{2} + y^{2}] = (x + 1)^{2} + y^{2}

after squaring. This simplifies to

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

, the equation of a fixed circle, and since A, B and C all obey the same ratio they all lie on it. The circumcentre of the triangle is the centre of this circle, namely

(\frac{5}{4}, 0)

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option

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JEE Main · 2009 · 12 Apr 2009 · Offline · Circle

Solution

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