Cor. 1.—If PI be produced to meet the circle again in L, PL is equal to PB. Cor. 2.—If two equal lines be drawn from P to either the convex or concave circumference, the diameter through P bisects the angle between them, and the parts of them intercepted by the circle are equal. Cor. 3.—If P be the common centre of circles whose radii are lines drawn from P to the circumference of HDE, then—1. The circle whose radius is the minimum line (PE) has contact of the first kind with ADE [Def. iv.]. 2. The circle whose radius is the maximum line (PA) has contact of the second kind. 3. A circle having any of the remaining lines (PF) as radius intersects HDE in two points (F, I). PROP. IX.—Theorem. A point (P) within a circle (ABC), from which more than two equal lines (PA, PB, PC, &c.) can be drawn to the circumference, is the centre. Dem.—If P be not the centre, let O be the centre. Join OP, and produce it to meet the circle in D and E; then DE is the diameter, and P is a point in it which is not the centre: therefore [vii.] only two equal lines can be drawn from P to the circumference; but three equal lines are drawn (hyp.), which is absurd. Hence P must be the centre. Or thus: Since the lines AP, BP are equal, the line bisecting the angle APB [vii. Cor. 1] must pass through the centre: in like manner the line bisecting the angle BPC must pass through the centre. Hence the point of intersection of these bisectors, that is, the point P, must be the centre. PROP. X.—Theorem. If two circles have more than two points common, they must coincide. Dem.—Let X be one of the circles; and if possible let another circle Y have three points, A, B, C, in common with X, without coinciding with it. Find P, the centre of X. Join PA, PB, PC. Then since P is the centre of X, the three lines PA, PB, PC are equal to one another. Again, since Y is a circle and P a point, from which three equal lines PA, PB, PC can be drawn to its circumference, P must be the centre of Y . Hence X and Y are concentric, which [v.] is impossible. Cor.—Two circles not coinciding cannot have more than two points common. Compare I., Axiom x., that two right lines not coinciding cannot have more than one point common. PROP. XI.—Theorem. If one circle (CPD) touch another circle (APB) internally at any point P, the line joining the centres must pass through that point. Dem.—Let O be the centre of APB. Join OP. I say the centre of the smaller circle is in the line OP. If not, let it be in any other position such as E. Join OE, EP, and produce OE through E to meet the circles in the points C, A. Now since E is a point in the diameter of the larger circle between the centre and A, EA is less than EP [vii. 2]; but EP is equal to EC (hyp.), being radii of the smaller circle. Hence EA is less than EC; which is impossible; consequently the centre of the smaller circle must be in the line OP. Let it be