Question

Consider two coplanar circles of radii R and r (R > r) with the same center. Let P be a fixed point on the smaller circle and B a variable point on the larger circle. The line BP meets the larger circle again at C. The perpendicular l to BP at P meets the smaller circle again at A. (If l is tangent to the circle at P then A = P.)
(i) Find the set of values of BC2 + CA2 + AB2.
(ii) Find the locus of the midpoint of BC.

Answer 1

Answer:

Let M be the midpoint of BC. Let PM = x. Let BC meet the small circle again at Q. Let O be the center of the circles. Since angle APQ = 90 degrees, AQ is a diameter of the small circle, so its length is 2r. Hence AP2 = 4r2 - 4x2. BM2 = R2 - OM2 = R2 - (r2 - x2). That is essentially all we need, because we now have: AB2 + AC2 + BC2 = (AP2 + (BM - x)2) + (AP2 + (BM + x)2) + 4BM2 = 2AP2 + 6BM2 + 2x2 = 2(4r2 - 4x2) + 6(R2 - r2 + x2) + 2x2 = 6R2 + 2r2 , which is independent of x.

(ii) M is the midpoint of BC and PQ since the circles have a common center. If we shrink the small circle by a factor 2 with P as center, then Q moves to M, and hence the locus of M is the circle diameter OP.

Step-by-step explanation:

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