I'm completely stuck on the second and third parts of this question. The first part I have done and detail below.

The circle S_{1}with centre C_{1}(a_{1},b_{1}) and radiusr_{1}touches externally the circle S_{2}with centre C_{2}(a_{2},b_{2}) and radiusr_{2}. The tangent at their common point passes through the origin.

(i) Show that (a_{1}^{2}-a_{2}^{2}) + (b_{1}^{2}-b_{2}^{2}) =r_{1}^{2}-r_{2}^{2}.

(ii) If, also, the other two tangents from the origin to S_{1}and S_{2}are perpendicular, prove that |a_{2}b_{1}-a_{1}b_{2}| = |a_{1}a_{2}+b_{1}b_{2}|.

(iii) Hence show that, if C_{1}remains fixed but S_{1}and S_{2}vary, then C_{2}lies on the curve (a_{1}^{2}-b_{1}^{2})(x^{2}-y^{2}) + 4a_{1}b_{1}xy= 0.

Part (i) I have solved by finding the equations of the two circles:

x^{2}+ y^{2}- 2a_{1}x - 2b_{1}y + a_{1}^{2}+ b_{1}^{2}= r_{1}^{2}

x^{2}+ y^{2}- 2a_{2}x - 2b_{2}y + a_{2}^{2}+ b_{2}^{2}= r_{2}^{2}

Then r_{1}^{2}- r_{2}^{2}= 2(a_{2}- a_{1})x + 2(b_{2}- b_{1})y + (a_{1}^{2}- a_{2}^{2}) + (b_{1}^{2}- b_{2}^{2}) [1]

The gradient of C_{1}C_{2}is (b_{2}- b_{1}) / (a_{2}- a_{1}) so the gradient of the common tangent is perpendicular to this, i.e. -(a_{2}- a_{1}) / (b_{2}- b_{1}) and the equation of this tangent through the origin is (b_{2}- b_{1})y = -(a_{2}- a_{1})x.

Substituting this in [1] gives (a_{1}^{2}-a_{2}^{2}) + (b_{1}^{2}-b_{2}^{2}) =r_{1}^{2}-r_{2}^{2}, as required.

For part (ii), I have established that dy/dx for S_{1}and S_{2}are given by -(x - a_{1})/(y - b_{1}) and -(x - a_{2})/(y - b_{2}), respectively. These are perpendicular, so

-(x - a_{1})/(y - b_{1}) = (y - b_{2})/(x - a_{2})

which leads to x^{2}+ y^{2}- (a_{1}+ a_{2})x - (b_{1}+ b_{2})y + a_{1}a_{2}+ b_{1}b_{2}= 0

And that's it - I'm completely stuck! Any help would be much appreciated. I have attached a diagram which I believe shows the situation.