Question

The value of k for (2k, 3k), (0, 0), (1, 0) and (0, 1) to be on the circle is

(1) \(\frac 2{13}\)

(2) \(\frac 5{13}\)

(3) \(\frac 1{13}\)

(4) \(\frac 3{13}\)

Answer 1

Correct option is (2) \(\frac 5{13}\)

Circle passing through (0, 0), (1, 0) and (0, 1) will be a circle having (1, 0) and (0, 1) as the end points of diameter.

C: (x – 1)x + y(y – 1) = 0

x² + y² – x – y = 0

Now (2k, 3k) lies on C.

4k² + 9k² – 2k – 3k = 0

13k² – 5k = 0

k(13k – 5) = 0

⇒ \(k = \frac 5{13}\)