Circle Question 4
Question 4 - 2024 (27 Jan Shift 1)
Four distinct points $(2 k, 3 k),(1,0),(0,1)$ and $(0,0)$ lie on a circle for $k$ equal to :
(1) $\frac{2}{13}$
(2) $\frac{3}{13}$
(3) $\frac{5}{13}$
(4) $\frac{1}{13}$
Show Answer
Answer (3)
Solution
$(2 k, 3 k)$ will lie on circle whose diameter is $AB$.
$(x-1)(x)+(y-1)(y)=0$
$x^{2}+y^{2}-x-y=0 \ldots$ (i)
Satisfy $(2 k, 3 k)$ in (i)
$(2 k)^{2}+(3 k)^{2}-2 k-3 k=0$
$13 k^{2}-5 k=0$
$k=0, k=\frac{5}{13}$
hence $k=\frac{5}{13}$
