Search any question & find its solution
Question:
Answered & Verified by Expert
If four distinct points $(2 k, 3 k),(2,0),(0,3)$ (0,0) lie on a circle, then
Options:
Solution:
2630 Upvotes
Verified Answer
The correct answer is:
$k=1$
Since, join of (2,0) and (0,3) subtends $90^{\circ}$ at (0,0)
$\Rightarrow$ It is a diameter.
$\therefore$ Equation is $(x-2)(x-0)+(y-0)(y-3)=0$
$x^{2}+y^{2}-2 x-3 y=0$
$(2 k, 3 k)$ lies on it $\Rightarrow \quad 4 k^{2}+9 k^{2}-4 k-9 k=0$
$\Rightarrow$
. $13 k^{2}=13 k$
$\Rightarrow$
$k=1$
since, $k \neq 0$ otherwise $(2 k, 3 k)$ will be (0,0)
$\Rightarrow$ It is a diameter.
$\therefore$ Equation is $(x-2)(x-0)+(y-0)(y-3)=0$
$x^{2}+y^{2}-2 x-3 y=0$
$(2 k, 3 k)$ lies on it $\Rightarrow \quad 4 k^{2}+9 k^{2}-4 k-9 k=0$
$\Rightarrow$
. $13 k^{2}=13 k$
$\Rightarrow$
$k=1$
since, $k \neq 0$ otherwise $(2 k, 3 k)$ will be (0,0)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.