The points (1, 0), (0, 1), (0, 0) and (2k, 3k),k ≠ 0 are concyclic if k =

Question

The points (1, 0), (0, 1), (0, 0) and (2k, 3k),k ≠ 0 are concyclic if k =  (A) - (5/13) (B)  (5/13) (C)  (1/5) (D) - (1/5)

Tardigrade Admin · Accepted Answer

The equation of the circle which passes through the points

(1, 0), (0, 1)

and

(0, 0)

is

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

Given that, the point

(2 k, 3 k)

is on the circle and form concyclic circle. Then, it satisfies the Eq. (i)

(2 k)^{2} + (3 k)^{2} - (2 k) - (3 k) = 0

\Rightarrow 4 k^{2} + 9 k^{2} - 5 k = 0

\Rightarrow 13 k^{2} - 5 k = 0

\Rightarrow k (13 k - 5) = 0

\Rightarrow k = 0

or

k = \frac{5}{13}

Hence,

k = \frac{5}{13}

Q. The points $(1, 0), (0, 1), (0, 0)$ and $(2 k, 3 k), k \neq = 0$ are concyclic if $k$ = _____