Show that the equation of a straight line meeting the circle$x^2+y^2=a^2$in two points at equal distances$'d'$from a point$(x_1,y_1)$

Question

Show that the equation of a straight line meeting the circle$x^2+y^2=a^2$in two points at equal distances$'d'$from a point$(x_1,y_1)$on its circumference is $xx_1+yy_1–a^2+d^2/2=0$. I tried to solve this problem by taking the centre origin and their making the line as chord of contact from the point given and making it another circle but could not solve it.

Answer 1

If $(h,k),(p,q)$ the two intersections $h^2+k^2=p^2+q^2=a^2$

$$d^2=(h-x_1)^2+(k-y_1)^2=h^2+k^2+x_1^2+y_1^2-2(hx_1+ky_1)$$

$$\iff2(hx_1+ky_1)+d^2-2a^2=0$$

Similarly, $$2(px_1+qy_1)+d^2-2a^2=0$$

So, the locus of the two intersections will be $$2(xx_1+yy_1)+d^2-2a^2=0$$