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

Mathematics · Circles

JEE Main 2024 — 27 January, Shift 1 — Question 12

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

$(2k, 3 k$ ) will lie on circle whose diameter is AB

$(x-1)(x)+(y-1)(y)=0$ $x^{2}+y^{2}-x-y=0$

Satisfy $(2 k, 3 k)$ in (i) $(2 \mathrm{k})^{2}+(3 \mathrm{k})^{2}-2 \mathrm{k}-3 \mathrm{k}=0$

$13 \mathrm{k}^{2}-5 \mathrm{k}=0$

$\mathrm{k}=0, \mathrm{k}=\frac{5}{13}$

hence $\mathrm{k}=\frac{5}{13}$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.