Question

If four points (0, 0), (1, 0), (0, 1), (2k, 3k) are concyclic, then k is 

(1) 4/13 

(2) 5/13 

(3) 7/13 

(4) 9/13

Answer 1

(2) 5/13 

Equation of circle is x(x – 1) + y(y – 1) = 0 

x² + y² – x – y = 0 

B(2k, 3k) 

⇒ 4k² + 9k² – 2k – 3k = 0 

⇒ 13k² = 5k

⇒ k = 0, 5/13

\(\therefore\) k = 5/13