A circle with diameter r is internally tangent to a circle with radius r at the point T. For some other point S on the larger circle, chord ST intersects the smaller circle at point X, and the tangents to a larger circle at S and T meet at point Y. Show that X, Y and the centre of the larger circle are collinear.
Let C be the center of the larger circle.
CX extended is a radius of the circle, so it is perpendicular to the chord ST, at its midpoint X.
The tangents SY and TY are the same length, so STY is an isosceles triangle, with altitude XY.
So, ST ⊥ XY, meaning C,X,Y are collinear.
To prove that X, Y, and the center of the larger circle are collinear, we need to show that the line XY passes through the center of the larger circle.
Let O be the center of the larger circle, and let M be the midpoint of ST.
First, let's draw the tangent to the larger circle at T. This tangent will be perpendicular to the line TM at point T.
Since TM is a chord of the smaller circle, it will be bisected by the line connecting the center of the smaller circle with the point X (let's call it Z). Therefore, ZM is perpendicular to TM.
Since TM is perpendicular to the tangent at T, and ZM is perpendicular to TM, we can conclude that ZM is parallel to the tangent at T.
Now, let's draw the tangent to the larger circle at S. This tangent will be perpendicular to the line SM at point S.
Since SM is a chord of the larger circle, it will be bisected by the line connecting the center of the larger circle with the point Y (let's call it N). Therefore, SN is perpendicular to SM.
Since SM is perpendicular to the tangent at S, and SN is perpendicular to SM, we can conclude that SN is parallel to the tangent at S.
Now, since ZM is parallel to the tangent at T and SN is parallel to the tangent at S, we can apply the converse of the Alternate Interior Angles theorem.
According to the converse of the Alternate Interior Angles theorem, if two lines are cut by a transversal in such a way that alternate interior angles are congruent, then the lines are parallel.
Therefore, ZM is parallel to SN.
Since ZM is parallel to SN and both lines pass through the center of the larger circle, O, we can conclude that Z, Y, and O are collinear.
To show that X, Y, and the center of the larger circle are collinear, we need to prove that XY passes through the center of the larger circle.
Let's denote the center of the larger circle as O, and the midpoint of segment ST as M. We'll start by proving that O, M, and X are collinear.
To do this, we'll consider the triangles OMT and OMX. Both triangles share the side OM, and we need to prove that the line XT is parallel to the line MO.
Since circle T is tangent to the larger circle at point T, the line XT is perpendicular to the radius OT at point T. Additionally, the line MO is also perpendicular to the radius OT at point M because M is the midpoint of the chord ST.
So we have two lines, XT and MO, that are each perpendicular to the same line, OT. Therefore, XT and MO must be parallel.
Now let's prove that Y lies on the line OX. To do this, we'll consider the triangles OYT and OXS.
Since the tangent to a circle is perpendicular to the radius of the circle at the point of tangency, we have that angles OTY and OSX are right angles.
Furthermore, we know that OT and OS are radii of the larger circle since they connect the center O to points T and S respectively. Therefore, OT = OS.
Using the two right angles and the shared length OT = OS, we can conclude that triangles OTY and OSX are congruent (by the theorem RHS - Right angle, Hypotenuse, Side). Therefore, angles OYT and OXS are equal.
Since angles OXS and OYT are equal, we can conclude that the lines OX and OY are parallel (because they both form equal angles with the line OT).
Now, when two parallel lines (OX and OM) are intersected by a transversal (XY), the corresponding angles are equal. Therefore, angles MOX and MOY are equal.
But we know that angle MOX is a right angle because MO is a perpendicular bisector of ST, and angle MOY is also a right angle because OY is tangent to the larger circle at point Y.
So we have two right angles that are equal, which means that angle MOX and angle MOY must be congruent.
Since angles MOX and MOY are congruent, and points X and Y lie on the line MO, we can conclude that X, Y, and the center of the larger circle (O) are collinear.