Theorem 3.1. If two circles meet at P and Q, then the magnitude of the angles between the circles is the same at P and Q.

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1 3 rthogonal circles Theorem 3.1. If two circles meet at and, then the magnitude of the angles between the circles is the same at and. roof. Referring to the figure on the right, we have A B AB (by SSS), so A B AB. Since the tangents to the circles at are perpendicular to the radii A and B, it follows that the angle between the tangents at is equal in measure to m( A B). Likewise, the angle between the tangents at is equal in measure to m( AB). A B Remark: The previous theorem means that to check the angle between two circles intersecting at and, we only need to check one of the angles. Note however, that the directions of the angles at and are opposite. Definition: Two intersecting circles and are said to be orthogonal if the angle between them is 90. We sometimes write to indicate orthogonality. Theorem 3.2. If two circles and are orthogonal, then (1) The tangents at each point of intersection pass through the centres of the other circle. (Figure (a) below). (2) Each circle is its own inverse with respect to the other. T (a) (b) roof. (1) This follows because a line through the point of tangency perpendicular to the tangent must pass through the centre of the circle. (2) Let be a point on the circle. Join to, the centre of, and let r be the radius of. Let be the other point where the ray meets. Let T be the point of intersection of the two circles so that T is the tangent to by part (1). By the power of the point with respect to we have = T 2 = r 2, showing that the inverse of any point on is another point on. 9

2 The preceding theorem has the following converse: Theorem 3.3. Two intersecting circles and are orthogonal if any one of the following statements is true. (1) The tangents to one circle at one point of intersection passes through the centre of the other circle (Figure (a) below). (2) ne of the circles passes through two distinct points that are inverses with respect to the other circle. r (a) (b) roof. (1) This implies that the two tangents at the point of intersection must be perpendicular. (2) Suppose that circle passes through and that are inverses with respect to. Let be the centre of and let be tangent to at (see Figure (b)). Then = 2 (by the power of with respect to ) = r 2 (since and are inverses). This implies that = r, so must be on as well as on. That is, is a point of intersection of and, and the tangent to at this point passes through the centre of. By part (1), the circles must be orthogonal. 10

3 The Arbelos Theorem (gilvy, p. 54; Eves, p. 133) Theorem 3.4. (The Arbelos Theorem a.k.a. appus Ancient Theorem).,, and R are three collinear points with C, D, and K 0 being semicircles on. R, and R. Let K 1, K 2,..., be circles touching C and D with K 1 touching K 0, K 2 touching K 1 and so on. Let h n be the distance of the centre of K n from R and let r n be the radius of K n. Then h n = 2nr n. D K n K 2 C h n K 1 R K 0 roof. m l D K n K 2 C K 1 R K 0 Let t be the tangential distance from to the circle K n, and apply I(, t 2 ). K n is orthogonal to the circle of inversion, so is its own inverse. C inverts into a line l. D inverts into a line m parallel to l. K 0 inverts into a semicircle K 0 tangent to l and m (because inversion preserves tangencies). K i inverts into a circle K i tangent to l and m. Then all of the K i have the same radius, namely r n, and the theorem follows. 11

4 Steiner s orism (gilvy, p ; Eves, p. 134, 135) Given a point outside a circle, a point of is said to be visible from if the segment meets only at. not visible from visible from (a) (b ) Figure (b) above shows in red the set of points that are visible from, namely the two tangent points and the points on the arc between the tangent points. In other words, a point of is visible from if and only if either is tangent to, or the tangent to at has and on opposite sides. Note also that if a line m is tangent to at, and if is on the same side of m as (but not on the line m), then is not visible from. Lemma 3.5. Suppose the line misses the circle. Then (1) There is a point visible from both and. (2) There is a point visible from but not from. (3) There is a point Z visible from but not from. The figure below illustrates how to find points and : Let m be a line parallel to and tangent to. is the point of tangency of m with. There are two lines from tangent to. Let l be the tangent line such that and are both on the same side of l. Then is the point where l is tangent to. l m 12

5 Lemma 3.6. Given two circles and, and given a point not on either circle, then there is a circle through orthogonal to both and. roof. S S Let be the inverse of with respect to. Let S be the inverse of with respect to. Then there is a unique circle through,, and S, and this circle must be orthogonal to both and by Theorem 3.3. Note: If,, and are collinear, then the orthogonal circle is a line. If,, and are not collinear, then the orthogonal circle is an true circle. Lemma 3.7. Let and be two non-intersecting circles with centres and,. Then we can find points and that are inverses to each other with respect to both and. roof. Let be any circle other than a line that is orthogonal to both and. We claim that the line intersects in two points. (Then the two points are and and they are inverses to each other with respect to both and.) case (i) case (ii) There are two cases to consider: (i) when the circles are exterior to each other and (ii) when one circle is inside the other. (i) Suppose, for a contradiction, that misses. Then there is a point Z on that is visible from both and. Since Z is visible from, it is inside or on. Since Z is visible from, it is inside or on. Then Z is inside or on both and, which contradicts the fact that and are exterior to each other. This proves case (i). (ii) Left as an exercise. 13

6 Definition. Let and be two non-intersecting circles. Let 1 be a circle tangent to both and. Let 2 be a circle tangent to 1,, and. Let 3 be a circle tangent to 2,, and. Continue in this fashion. If at some point k is tangent to 1 we say that 1, 2,..., k is a Steiner chain of k circles. Remark: Given two circles and, there is no guarantee that a Steiner chain exists for and. Theorem 3.8. (Steiner s orism.) Suppose that the two non-intersecting circles and have a Steiner chain of k circles. Then, any circle tangent to and is a member of some Steiner chain of k circles. To prove this theorem, we need the following: Lemma 3.9. Given two non-intersecting circles and (that are not concentric), there is an inversion that transforms them into concentric circles. roof. sing Lemma 3.6 we can find two circles and simultaneously orthogonal to both and. These two circles intersect at points and referred to in Lemma 3.7. ' ' ' ' ' erform the inversion I(, r 2 ) for some radius r. Then transforms to (a straight line through and not through ). transforms to (a straight line through and not through ). transforms to a circle, and, (because orthogonality is preserved). transforms to a circle, and,. Since the circle is orthogonal to the line, then must be centred at some point of. Similarly, must be centred at some point of. Thus, is centred at. By the same argument, must also be centred at. 14

7 roof of Steiner s orism. ' Invert and into concentric circles. The inversion preserves the Steiner chain of k circles. ' ' sing the same inversion, transform into a circle '. ' ' The circle ' is obviously part of a Steiner chain of k circles, so by the reverse inversion must also be part of a Steiner chain of k circles. 15