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 A B is :

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Aicharya · Accepted Answer

Since, circle passing through origin intersect the coordinate axes at $A & B$ , hence $A B$ must be diameter and $A B = 2 R$ .

Now, let foot of the perpendicular from origin upon $A B$ be $P (h, k)$ .

Slope of line $O P = \frac{k - 0}{h - 0} = \frac{k}{h}$

Since, line $A B ⊥ O P \Rightarrow$ slope of $A B = - \frac{h}{k}$

Thus, equation of line $A B$ is $y - k = \frac{- h}{k} (x - h)$

For co-ordinates of $A$ , put $y = 0 \Rightarrow 0 - k = \frac{- h}{k} (x - h) \Rightarrow x = \frac{h^{2} + k^{2}}{h} \Rightarrow A (\frac{h^{2} + k^{2}}{h}, 0)$ .

For co-ordinates of $B$ , put $x = 0 \Rightarrow y - k = \frac{- h}{k} (0 - h) \Rightarrow y = \frac{h^{2} + k^{2}}{k} \Rightarrow B (0, \frac{h^{2} + k^{2}}{k})$ .

Now, given $A B = 2 R$

Applying distance formula,

$\Rightarrow \sqrt{{(\frac{h^{2} + k^{2}}{h} - 0)}^{2} + {(0 - \frac{h^{2} + k^{2}}{k})}^{2}} = 2 R$

$\Rightarrow {(h^{2} + k^{2})}^{2} (\frac{1}{h^{2}} + \frac{1}{k^{2}}) = 4 R^{2} \Rightarrow {(h^{2} + k^{2})}^{3} = 4 R^{2} h^{2} k^{2}$

Hence, locus is ${(x^{2} + y^{2})}^{3} = 4 R^{2} x^{2} y^{2}$

If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B , then the…

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