Question

Show that the equation of a straight line meeting the circle

x^{2} + y^{2} = a^{2}

in two points at equal distance '

d

' from a point

(x_{1}, y_{1})

on its circumference is

x x_{1} + y y_{1} - a^{2} + \frac{d^{2}}{2} = 0

.

Accepted Answer

The equation of the circle is $x^{2} + y^{2} = a^{2}$ .

Let the points the line meets circle be $A (a \cos θ_{1}, a \sin θ_{1})$ and $B (a \cos θ_{2}, a \sin θ_{2})$ .

Given these points are at a distance $d$ from $(x_{1}, y_{1})$

$\Rightarrow {(x_{1} - a \cos θ_{1})}^{2} + {(y_{1} - a \sin θ_{1})}^{2} = {(x_{1} - a \cos θ_{2})}^{2} + {(y_{2} - a \sin θ_{2})}^{2} = d^{2} . . . (1)$

$\Rightarrow x_{1}^{2} - 2 a x_{1} \cos θ_{1} + a^{2} {\cos^{2} θ}_{1} + y_{1}^{2} + a^{2} {\sin^{2} θ}_{1} - 2 a y_{1} \sin θ_{1} = x_{1}^{2} - 2 a x_{1} \cos θ_{2} + a^{2} \cos^{2} θ_{2} + y_{1}^{2} - 2 a y_{1} \sin θ_{2} + a^{2} \sin^{2} θ_{2}$

$\Rightarrow x_{1} \cos θ_{1} + y_{1} \sin θ_{1} = y_{1} \sin θ_{2} + x_{1} \cos θ_{2}$

$\Rightarrow \frac{\sin θ_{2} - \sin θ_{1}}{\cos θ_{2} - \cos θ_{1}} = \frac{- x_{1}}{y_{1}} . . . . (2)$

The equation of the line joining A and B is $(y - a \sin θ_{2}) = \frac{\sin θ_{2} - \sin θ_{1}}{\cos θ_{2} - \cos θ_{1}} (x - a \cos θ_{2})$

$\Rightarrow (y - a \sin θ_{2}) = \frac{- x_{1}}{y_{1}} (x - a \cos θ_{2})$

$\Rightarrow y_{1} y + x x_{1} - y_{1} a \sin θ_{2} - a x_{1} \cos θ_{2} = 0$

$\Rightarrow x x_{1} + y y_{1} + (\frac{x_{1}^{2} - 2 a x_{1} \cos θ_{2} + a^{2} \cos^{2} θ_{2}}{2}) + (\frac{y_{1}^{2} - 2 a y_{1} \sin θ_{2} + a^{2} \sin^{2} θ_{2}}{2}) - (\frac{x_{1}^{2} + y_{1}^{2}}{2}) - \frac{a^{2}}{2} = 0$

$\Rightarrow x x_{1} + y y_{1} - a^{2} + \frac{d^{2}}{2} = 0$

Hence, proved.

EXERCISE-4

Embibe Experts Mathematics Solutions for EXERCISE-4

Questions from EXERCISE-4 with Hints & Solutions

Learn and Prepare for any exam you want !