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 $x{x}_1+y{y}_1-a^2+\left(\frac{d^2}{2}\right)=0$.

Answer 1

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

Let the points the line meets circle be,

$A(acos{\theta }_1,asin{\theta }_1)$ and $B(acos{\theta }_2,asin{\theta }_2)$

Given these points are at distanc 'd' from $(x_1,y_1)$

$\Rightarrow (x_1-acos{\theta }_1{)}^2+(y_1-asin{\theta }_1{)}^2=(x_1-acos{\theta }_2{)}^2+(y_2-asin{\theta }_2{)}^2=d^2...\left(1\right)$

$\Rightarrow x_1^2-2a{x}_{1}cos{\theta }_1+a^{2}co{s}^2{\theta }_1^2+y_1^2+a^{2}si{n}^2{\theta }_1^2$

$-2a{y}_{1}sin{\theta }_1=x_1^2-2a{x}_{1}cos{\theta }_2+a^{2}co{s}^2{\theta }_2+y_1^2$

$-2a{y}_{1}sin{\theta }_2+a^{2}si{n}^2{\theta }_2$

$x_{1}cos{\theta }_1+y_{1}sin{\theta }_1=y_{1}sin{\theta }_2+x_{1}cos{\theta }_2$

$\Rightarrow \frac{sin{\theta }_2-sin{\theta }_1}{cos{\theta }_2-cos{\theta }_1}=\frac{-x_1}{y_1}...\left(2\right)$

The equation of the line joining A and B is

$(y-asi{n}_2)=\frac{asin{\theta }_2-asin{\theta }_1}{acos{\theta }_2-acos{\theta }_1}(x-acos{\theta }_2)$

$\Rightarrow (y-asin{\theta }_2)=\frac{-x_1}{y_1}(x-acos{\theta }_2)$

$\Rightarrow y_{1}y+x{x}_1-y_{1}asin{\theta }_2-a{x}_{1}cos{\theta }_2=0$

$\Rightarrow x{x}_1+y{y}_1+y{y}_1+\left(\frac{x_1^2-2a{x}_{1}cos{\theta }_2+a^{2}co{s}^2{\theta }_2}{2}\right)$

$+\left(\frac{y_1^2-2a{y}_{1}sin{\theta }_2+a^{2}si{n}^2{\theta }_2}{2}\right))-(\frac{x_1^2+y_1^2}{2})-\frac{a^2}{2}=0$

$\Rightarrow x{x}_1+y{y}_1+\frac{d^2}{2}-a^2=0$