Question P.412: We are given an angle AOB and a point P. 1°. Find a point M ......
We are given an angle \widehat{{A O B}} and a point P.
{1}^{\circ}. Find a point M on the side OA such that the two circles C,C^{\prime} tangent to OB and passing through the points M, P intersect at a given angle; .
{2}^{\circ}. Study the variation of the angle between C and {{C}}^{\prime} as M moves on OA;
{3}^{\circ}. Let Q,\ Q^{\prime} be the points (other than M) where these circles intersect the side OA. Show that circle through P, Q, and Q^{\prime} is tangent to a fixed line as M moves on OA (this reduces to the preceding exercise).
We take the original figure, shown as figure t412a, and invert it around pole P by any power. To make things simpler, we choose the power to be OP², so that the circle of inversion has radius OP. The result of the inversion is shown in figure t412b, where corresponding elements are labeled with the subscript 1. The points O, P are their own images, so we will sometimes refer to O as {{O}}_{1} and to P as P_{1}.
{1}^{\circ}. Circles C,\ C^{\prime} invert into two lines C_{1},\ C_{1}^{\prime}, both tangent to circle O_{1}B_{1}P_{1} and passing through point \mathrm{M}_{1} on circle O_{1}A_{1}P_{1}. The two angles at which circles C \text{ and } C^{\prime} meet are equal to the two angles formed by lines C_{1},\ C_{1}^{\prime}. In particular, the angle lying outside both circles, in the region of the plane containing line OB, is equal to the angle in the region of the plane containing circle O B_{1}P. So we have reduced the problem to that of finding a point on circle O A_{1}P at which the tangents to circle O B_{1}P meet at a given angle.
The locus of points whose tangents to a given circle meet at a given angle is itself a circle, concentric to the given circle. Therefore point \mathrm{M}_{1} is the intersection of circle O A_{1}P and some circle S_{1}, the locus of points such that the tangents from those points to circle O B_{1}P meet at an angle equal to that at which circles C,\ C^{\prime} meet. Then M is the intersection of line OA with circle S, the inverse image of S_{1}.
{2}^{\circ}. As point \mathrm{M}_{1} moves along arc \overset{︵}{OA_1P}, point M moves along side OA of angle {\widehat{O A B}}. the angle between C and C^{\prime} decreases from 180^{\circ} to some minimum determined by the greatest distance from \mathrm{M}_{1} to the center of circle O B_{1}P, then increases to 180^{\circ}.
{3}^{\circ}. Circle P Q Q^{\prime} inverts into line Q_{1}Q_{1}^{\prime}, which joins the second points of in-tersection of tangents from \mathrm{M}_{1} to circle O B_{1}P with circle { O}A_{1}{P}. But, by exercise 411, line Q_1 Q_1^{\prime} remains tangent to a certain circle, which has a common radical axis with circles O A_{1}P,\;O B_{1}P.Therefore this circle passes through points O and P. Thus, circle P Q Q^{\prime} which is the inversion of line Q_{1}Q_{1}^{\prime}, must remain tangent to some line through point O.
