Coordinate geometry problem

[jimmyjunk19]

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

The circle S₁ with centre C₁(a₁, b₁) and radius r₁ touches externally the circle S₂ with centre C₂(a₂, b₂) and radius r₂. The tangent at their common point passes through the origin.

(i) Show that (a₁² - a₂²) + (b₁² - b₂²) = r₁² - r₂².

(ii) If, also, the other two tangents from the origin to S₁ and S₂ are perpendicular, prove that |a₂b₁ - a₁b₂| = |a₁a₂ + b₁b₂|.

(iii) Hence show that, if C₁ remains fixed but S₁ and S₂ vary, then C₂ lies on the curve (a₁² - b₁²)(x² - y²) + 4a₁b₁xy = 0.

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

x² + y² - 2a₁x - 2b₁y + a₁² + b₁² = r₁²
x² + y² - 2a₂x - 2b₂y + a₂² + b₂² = r₂²

Then r₁² - r₂² = 2(a₂ - a₁)x + 2(b₂ - b₁)y + (a₁² - a₂²) + (b₁² - b₂²) [1]

The gradient of C₁C₂ is (b₂ - b₁) / (a₂ - a₁) so the gradient of the common tangent is perpendicular to this, i.e. -(a₂ - a₁) / (b₂ - b₁) and the equation of this tangent through the origin is (b₂ - b₁)y = -(a₂ - a₁)x.

Substituting this in [1] gives (a₁² - a₂²) + (b₁² - b₂²) = r₁² - r₂², as required.


For part (ii), I have established that dy/dx for S₁ and S₂ are given by -(x - a₁)/(y - b₁) and -(x - a₂)/(y - b₂), respectively. These are perpendicular, so

-(x - a₁)/(y - b₁) = (y - b₂)/(x - a₂)

which leads to x² + y² - (a₁ + a₂)x - (b₁ + b₂)y + a₁a₂ + b₁b₂ = 0

And that's it - I'm completely stuck! Any help would be much appreciated.