Solution

The equation of the circle is $$ x^{2}+y^{2} = a^{2} $$
Let the points the line meets circle be,
$$\displaystyle A (a\,cos\,\theta _{1},a\,sin\,\theta _{1}) $$ and $$ B(a\,cos\,\theta _{2},a\,sin\,\theta _{2}) $$
Given these points are at distanc 'd' from $$ (x_{1},y_{1}) $$
$$\displaystyle \Rightarrow (x_{1}-a\,cos\,\theta _{1})^{2}+(y_{1}-a\,sin\,\theta _{1})^{2} = (x_{1}-a\,cos\,\theta _{2})^{2} +(y_{2}-a\,sin\,\theta _{2})^{2} = d^{2} ...(1) $$
$$\displaystyle \Rightarrow x_{1}^{2}-2a\,x_{1}\,cos\,\theta _{1}+a^{2}\,cos^{2}\theta _{1}^{2}+y_{1}^{2}+a^{2}\,sin^{2}\theta _{1}^{2} $$
$$\displaystyle -2ay_{1}\,sin\theta _{1} = x_{1}^{2}-2ax_{1}\,cos\theta _{2}+a^{2}\,cos^{2}\,\theta _{2}+y_{1}^{2} $$
$$ -2ay_{1}\,sin\theta _{2}+a^{2}\,sin^{2}\,\theta _{2} $$
$$ x_{1}\,cos\,\theta _{1}+y_{1}\,sin\theta _{1} = y_{1}\,sin\theta _{2}+x_{1}\,cos\theta _{2} $$
$$\displaystyle \Rightarrow \frac{sin\theta _{2}-sin\theta _{1}}{cos\theta _{2}-cos\theta _{1}} = \frac{-x_{1}}{y_{1}} ... (2) $$
The equation of the line joining A and B is
$$\displaystyle (y-a\,sin_{2}) = \frac{a\,sin\theta _{2}-a\,sin\theta _{1}}{a\,cos\theta _{2}-a\,cos\theta _{1}} (x-a\,cos\theta _{2}) $$
$$\displaystyle \Rightarrow (y-a\,sin\,\theta _{2}) = \frac{-x_{1}}{y_{1}}(x-a\,cos\,\theta _{2}) $$
$$\displaystyle \Rightarrow y_{1}y+xx_{1}-y_{1}\,a\,sin\,\theta _{2}-a\,x_{1}\,cos\theta _{2} = 0 $$
$$\displaystyle \Rightarrow xx_{1}+yy_{1}+yy_{1}+(\frac{x_{1}^{2}-2ax_{1}\,cos\,\theta _{2}+a^{2}\,cos^{2}\theta _{2}}{2})$$
$$ \displaystyle+(\frac{y_{1}^{2}-2ay_{1}\,sin\theta _{2}+a^{2}\,sin^{2}\theta _{2}}{2}))-(\frac{x_{1}^{2}+y_{1}^{2}}{2}) -\frac{a^{2}}{2} = 0 $$
$$\displaystyle \Rightarrow xx_{1}+yy_{1}+\frac{d^{2}}{2}-a^{2} = 0 $$