First slide

Definition of a circle

Question

The tangent at any point to the circle $x^{2} + y^{2} = 16$ meets the coordinate axes at A and B. If the lines drawn parallel to axes through A and B meet at ‘P’ Then locus of ‘P’ is

Easy

A

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = 4$
B

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{8}$
C

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{16}$
D

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = 16$

Solution

Given circle is $x^{2} + y^{2} = 16$
$C = (0, 0) . r = 4$
$l e t P (x_{1}, y_{1})$ be the point on locus
$\Rightarrow A (x_{1}, 0) B (0, y_{1})$
$∴ e q . o f \vec{A B} i s \frac{x}{x_{1}} + \frac{y}{y_{1}} = 1$ …………….(1)
Since equation $(1)$ touches the circle
$r = d \Rightarrow 4 = \frac{1}{\sqrt{\frac{1}{x_{1}^{2}} + \frac{1}{y_{1}^{2}}}}$ $\Rightarrow \sqrt{\frac{1}{x_{1}^{2}} + \frac{1}{y_{1}^{2}}} = \frac{1}{4}$
S.O.B.S
$\Rightarrow \frac{1}{x_{1}^{2}} + \frac{1}{y_{1}^{2}} = \frac{1}{16}$
$∴$ locus is $\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{16}$

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