The circle C has diameter AB. The tangent at B is T. For each point M (not equal to A) on C there is a circle C' which touches T and touches C at M. Find the point at which C' touches T and find the locus of the center of C' as M varies. Show that there is a circle orthogonal to all the circles C'.
Answer
C' touches T at the intersection of T and the line AM
the locus of the center is a parabola vertex B
the circle center A radius AB is orthogonal to all circles C'
Solution
Let O be the center of C. Let the line AM meet T at N. Let the perpendicular to T at N meet the line OM at O'. Then ∠O'NM = ∠MAB (O'N parallel to AB, because both perpendicular to T) = ∠OMA (OM = OA) = ∠O'MN. So O'M = O'N. Hence O' is the center of C'.
Take B to be the origin and A to be the point (2a,0), so O is (a,0) and C has radius a. If O' is (x,y), then we require that O'O = x+a or (x-a)2+y2 = (x+a)2, or y2 = 4ax, which is a parabola with vertex B and axis the x-axis.
Triangles AMB, ABN are similar (∠AMB = ∠ABN = 90o), so AM/AB = AB/AN and hence AM·AN = AB2. Now consider the circle center A radius AB. It must meet the circle C', because it contains the point M. Suppose it meets at X. Then AX2 = AB2 = AM·AN, so AX is tangent to C' and hence the circles are orthogonal.
© John Scholes
jscholes@kalva.demon.co.uk
4 February 2004
Last corrected/updated 4 Feb 04