Four distinct points (2 k, 3 k),(1,0),(0,1) and (0,0) lie on a circle for k equal to :

Question

Four distinct points $(2 k, 3 k),(1,0),(0,1)$ and $(0,0)$ lie on a circle for $k$ equal to :

Accepted Answer

$(2 k, 3 k)$ will lie on circle whose diameter is $A B$.

$$\begin{aligned}
& (x-1)(x)+(y-1)(y)=0 \
& x^2+y^2-x-y=0 \quad \text{.... (i)}
\end{aligned}$$

Satisfy (2k, 3k) in (i)

$$\begin{aligned}
& (2 k)^2+(3 k)^2-2 k-3 k=0 \
& 13 k^2-5 k=0 \
& k=0, k=\frac{5}{13} \
& \text { hence } k=\frac{5}{13}
\end{aligned}$$

Solutions · 1