JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 6)
If a circle of radius R passes through the origin O and intersects the coordinates axes at A and B, then the
locus of the foot of perpendicular from O on AB is :
(x2 + y2)2 = 4R2x2y2
(x2 + y2) (x + y) = R2xy
(x2 + y2)2 = 4Rx2y2
(x2 + y2)3 = 4R2x2y2
Explanation
_12th_January_Evening_Slot_en_6_1.png)
Slope of AB = $${{ - h} \over k}$$
Equation of AB is hx + ky = h2 + k2
A $$\left( {{{{h^2} + {k^2}} \over h},0} \right),B\left( {0,{{{h^2} + {k^2}} \over k}} \right)$$
AB = 2R
$$ \Rightarrow $$ (h2 + k2)3 = 4R2h2k2
$$ \Rightarrow $$ (x2 + y2)3 = 4R2x2y2
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