# 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$(x_1,y_1)$

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.

• You should really try to come up with a better title than a fragment of the first sentence of your question. – amd Oct 19 '18 at 21:38

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 $$2(xx_1+yy_1)+d^2-2a^2=0$$

• Why did you do $h^2+k^2=p^2+q^2$?? – priyanka kumari Oct 19 '18 at 11:21
• @priyanka, Both $(h,k);(p,q)$ lie on the same given circle – lab bhattacharjee Oct 19 '18 at 11:37
• Yes!! I realised it now thank you. – priyanka kumari Oct 19 '18 at 11:38