JEE PYQ: Circle Question 41
Question 41 - 2019 (12 Jan Shift 1)
If a circle of radius $R$ passes through the origin $O$ and intersects the coordinate axes at $A$ and $B$, then the locus of the foot of perpendicular from $O$ on $AB$ is:
(1) $(x^2 + y^2)^2 = 4R^2 x^2 y^2$
(2) $(x^2 + y^2)^3 = 4R^2 x^2 y^2$
(3) $(x^2 + y^2)^2 = 4Rx^2 y^2$
(4) $(x^2 + y^2)(x + y) = R^2 xy$
Show Answer
Answer: (2)
Solution
$AB$ is the diameter: $A = (\frac{h^2+k^2}{h}, 0)$, $B = (0, \frac{h^2+k^2}{k})$. With $AB = 2R$: $\frac{(h^2+k^2)^2}{h^2} + \frac{(h^2+k^2)^2}{k^2} = 4R^2$. So $(h^2+k^2)^3 = 4R^2 h^2 k^2$.
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