Question

If four distinct points (2k,3k),(2,0),(0,3),(0,0) lie on a circle, then
A. k<0
B. 0C. k=1
D. k>1

Answer 1

Hint – In order to solve this problem we have to use the concept that the endpoints of diameter subtends right angle at the circle and the circle whose endpoints of diameter are given as (a,b) and (c,d) then its equation will be (x - a)(x - c)+(y - b)(y - d) = 0. Using these concepts you can find the location of k.

Complete step-by-step solution -

The diagram of the circle can be drawn as :-

Since, join of (2,0) and (0,3) subtends right angle at (0,0)
 So, it is a diameter.
∴ Equation is (x−2)(x−0)+(y−0)(y−3)=0
${{\text{x}}^{\text{2}}}{\text{ - 2x + }}{{\text{y}}^{\text{2}}}{\text{ - 3y = 0}}$
It is given that (2k,3k) lies on the circle so we put (2k,3k) in the equation of circle we get the equation as:

$ \Rightarrow $${{\text{(2k)}}^{\text{2}}}{\text{ - 2(2k) + 3}}{{\text{k}}^{\text{2}}}{\text{ - 3(3k) = 0}}$
$ \Rightarrow $${\text{4}}{{\text{k}}^{\text{2}}}{\text{ + 9}}{{\text{k}}^{\text{2}}}{\text{ - 4k - 9k = 0 }}$
$ \Rightarrow $${\text{13}}{{\text{k}}^{\text{2}}}{\text{ - 13k = 0}}$
$ \Rightarrow $${\text{k(k - 1) = 0}}$
$ \Rightarrow $k=1, k=0
Since, ${\text{k}} \ne {\text{0}}$ otherwise (2k,3k) will be (0,0).
Therefore, the value of k is 1.
So, the correct option is C.

Note – Whenever you face such a type of circle you have to use the concept of circles. Here we have used the concepts that ends of diameter subtends right angle on the circle and the equation of circle whose endpoints of one of the diameter is given [say (a,b) and (c,d)] can be written as (x-a)(x-b)+(y-c)(y-d)=0. Using these concepts will take you to the right answer.