Question
Four distinct points $$\left( {2k,\,3k} \right),\,\left( {1,\,0} \right),\,\left( {0,\,1} \right)$$ and $$\left( {0,\,0} \right)$$ lie on a circle for :
A.
only one value of $$k$$
B.
$$0 < k < 1$$
C.
$$k < 0$$
D.
all integral values of $$k$$
Answer :
only one value of $$k$$
Solution :
The equation of the circle through $$\left( {1,\,0} \right),\,\left( {0,\,1} \right)$$ and $$\left( {0,\,0} \right)$$ is $${x^2} + {y^2} - x - y = 0$$
It passes through $$\left( {2k,\,3k} \right)$$
$$\eqalign{
& {\text{So, }}4{k^2} + 9{k^2} - 2k - 3k = 0 \cr
& \Rightarrow 13{k^2} - 5k = 0 \cr
& \Rightarrow k\left( {13k - 5} \right) = 0 \cr
& \Rightarrow k = 0{\text{ or }}k = \frac{5}{{13}} \cr} $$
But $$k \ne 0$$ [$$\because $$ all the four points are distinct]
$$\therefore k = \frac{5}{{13}}$$