JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Circle

Four distinct points

(2 k, 3 k)

(1, 0)

(0, 1)

and

(0, 0)

lie on a circle for k equal to :

(A) $\frac{2}{13}$
(B) $\frac{3}{13}$
(C) $\frac{5}{13}$ ✓ Correct
(D) $\frac{1}{13}$

Answer: (C)

Solution

The circle through

(0, 0), (1, 0), (0, 1)

x^{2} + y^{2} - x - y = 0

(from

x^{2} + y^{2} + D x + E y + F = 0

with

F = 0

D = E = - 1

).

Substituting

(2 k, 3 k)

4 k^{2} + 9 k^{2} - 2 k - 3 k = 0 \Rightarrow 13 k^{2} - 5 k = 0 \Rightarrow k (13 k - 5) = 0

Since the four points are distinct,

k \neq = 0

, so

k = \frac{5}{13}

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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