Question

Prove that the locus of a point which moves such that the difference of the squares of the lengths of tangents drawn from it to two given circles is constant is a line parallel to the radical axis of the given circles.

Answer 1

Length of the given tangent =$\sqrt{(}{x}^2+y^2+2g_{1}x+2f_{1}y+c_1)$

Length of the another tangent= $\sqrt{(}{x}^2+y^2+2g_{2}x+2f_{2}y+c_2)$

squaring of the length of tangents =$(x^2+y^2+2g_{1}x+2f_{1}y+c_1)$$,(x^2+y^2+2g_{2}x+2f_{2}y+c_2)$

subtracting 1 from 2

$(x^2+y^2+2g_{1}x+2f_{1}y+c_1)$$-(x^2+y^2+2g_{2}x+2f_{2}y+c_2)=$

$2(g_1-g_2)x+2(f_1-f_2)y+c_1-c_2=0$

hence the given locus is a line parallel to the radical axis of the given circles