1. Coordinate geometry problem

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. I have attached a diagram which I believe shows the situation.

2. Re: Coordinate geometry problem

Good effort thus far!

By doing some geometry, you should see that angle formed between vectors $(a_1, b_1)$ and $(a_2, b_2)$ is 45 degrees.

Next, use the geometric definition of the dot product:

$(a_1,b_1)\cdot (a_2,b_2) = (a_1^2+b_1^2)^{\tfrac{1}{2}} (a_2^2+b_2^2)^{\tfrac{1}{2}}\cos{45^{\circ}}$

Can you finish part (ii) now?

-Andy

3. Re: Coordinate geometry problem

Many thanks, I have now solved part (ii), although maybe not in the way you envisioned. The vital clue was that the angle between OA and OB is 450.

I noted that, since triangles AOD and AOE are similar, and triangles BOC and BOE are similar, then if the angle AOD is p and the angle BOC is q, then we have 2(p + q) = 90o, so the angle between OA and OB is p + q = 45o.

The angle between the lines OA and OB is given by

tan(p + q) = (b1/a1 - b2/a2)/(1+(b1b2/a1a2) = 1

so

|a2b1 - a1b2| = |a1a2 + b1b2|

Now on to try part (iii)

Nice work!

5. Re: Coordinate geometry problem

I've been following this thread with interest for some time and have made a few attempts at it. My last attempt was along the lines of post #3, but can someone please explain the reasoning to this line - which appears in post #3:

The angle between the lines OA and OB is given by
tan(p + q) = (b1/a1 - b2/a2)/(1+(b1b2/a1a2) = 1

[I can see that tan(p + q) = tan 45° = 1]

The line in question appears similar, but not same, as tan(A + B) = (tanA + tanB) / (1-tanA*tanB)

(From that quoted line, I follow the remaining reasoning to the conclusion).

Many thanks,
Al. / June 16th.

6. Re: Coordinate geometry problem

That comes from the formula for the angle between two lines. You are on the right track with a tangent formula.

Consider two lines with gradients m1 (= tan t1, say) and m2 (= tan t2, say). The angle between the lines, p, is given by

p = t1 - t2

Therefore tan p = tan(t1 - t2) = (tan t1 - tan t2)/(1 + tan t1 tan t2)

i.e.

tan p = (m1 - m2) / (1 + m1m2)

In the case of this question, m1 = b1/a1 and m2 = b2/a2, which leads to the derived expression.

7. Re: Coordinate geometry problem

Thank you.

Al. / Skywave