Question

Two circles with centres C and D, intersect each other at points P and Q. If APB is parallel to CD, prove that AB = 2·CD

Answer 1

Given: Two circles with centres C and D intersect at points P and Q. The line APB i.e., the straight line through A, P and B is parallel to CD.

To Prove: AB = 2·CD

Proof [Step-wise]:

1. Set up coordinates (convenient choice).

Place C at (0, 0) and D at (d, 0), so CD = d.

Because APB || CD, the line APB is horizontal; let its equation be y = h(h > 0).

Let P be the common intersection point of the two circles with coordinates P(x₀, h).

2. Let r₁ = CP and r₂ = DP be the radii of the left and right circles, respectively.

Because P lies on both circles.

x₀² + h² = r₁²   ...(1) 

(x₀ − d)² + h² = r₂²   ...(2)

3. Subtract (2) from (1) if needed this relation will be used implicitly below.

More directly, note the horizontal line y = h meets the left circle at two x-values `±sqrt(r_1^2 - h^2)`.

One of these x-values is x₀ the x-coordinate of P; the other is the point A on the left circle.

So, `x_A = -sqrt(r_1^2 - h^2)`

`x_P = +sqrt(r_1^2 - h^2)`   ...(3)

Similarly, the same horizontal line meets the right circle at `x = d ± sqrt(r_2^2 - h^2)`.

One of these is P,

So, `x_P = d - sqrt(r_2^2 - h^2)`

`x_B = d + sqrt(r_2^2 - h^2)`   ...(4)

Equating the two expressions for x_P from (3) and (4) gives

`sqrt(r_1^2 - h^2) = d - sqrt(r_2^2 - h^2)`   ...(5)

4. Compute AB as the difference of x-coordinates of B and A:

AB = x_B – x_A 

= `[d + sqrt(r_2^2 - h^2)] - [-sqrt(r_1^2 - h^2)]`

= `d + sqrt(r_2^2 - h^2) + sqrt(r_1^2 - h^2)`

Substitute (5) i.e., `sqrt(r_1^2 - h^2) = d - sqrt(r_2^2 - h^2)` into the expression:

`AB = d + sqrt(r_2^2 - h^2) + (d - sqrt(r_2^2 - h^2))`

AB = 2d

5. Since d = CD, we obtain AB = 2·CD.