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(Theorem 1) Is Proved Rad HAZU Volume 503 (2009), 41–54 ABOUT ONE RELATION CONCERNING TWO CIRCLES MIRKO RADIC´ AND ZORAN KALIMAN Abstract. This article can be considered as an appendix to the article [1]. Here the article [1] is extended to the cases when one circle is outside of the other and when circles are intersecting. 1. Preliminaries In [1] the following theorem (Theorem 1) is proved: Let C1 and C2 be any given two circles such that C1 is inside of the C2 and let A1 , A2 , A3 be any given three different points on C2 such that there are points T1 and T2 on C1 with properties |A1A2| = t1 +t2, |A2A3| = t2 +t3, (1) where t1 = |A1T1|, t2 = |T1A2|, t3 = |T2A3|.Then 2rR |A A | =(t +t ) , (2) 1 3 1 3 R2 − d2 where r = radius of C1 , R = radius of C2 , d = |IO|, I is the center of C1 , O is center of C2 .(SeeFigure1.) In short about the proof of this theorem. First the following lemma is proved. It t1 is given then t2 can be calculated using the expression √ t (R2 − d2) ± D (t ) = 1 1 (3a) 2 1,2 2 + 2 r t1 where = 2( 2 − 2)2 +( 2 + 2) 2 2 − 2 2 − ( 2 + 2 − 2)2 . D1 t1 R d r t1 4R d r t1 R d r (3b) The values (t2)1,2 given by (3) are solutions of the equation ( 2 + 2) 2 − ( 2 − 2)+ 2 2 − 2 2 +( 2 + 2 − 2)2 = . r t1 t2 2t1t2 R d r t1 4R d R d r 0 (4) Mathematics subject classification (2000): 51M04. Keywords and phrases: two circles; relation. (Received March 19, 2006) (Accepted May 29, 2007) 41 Rad Hrvat. akad. znan. umjet. 503. Matematickeˇ znanosti 16 (2009), str. 41–54 Mirko Radic´ and Zoran Kaliman About one relation concerning two circles Figure 1 This equation is obtained from the equations 2 2 2 (t1 +t2) =(x1 − x2) +(y1 − y2) , (5a) 2 =( − )2 + 2 − 2, 2 =( − )2 + 2 − 2 t1 x1 d y1 r t2 x2 d y2 r (5b) 2 + 2 = 2 + 2 = 2 ( ) using relations x1 y1 x2 y2 R . (See Figure 2.) The length t2 1 in Figure 2 is denoted by t2 . Figure 2 42 Rad Hrvat. akad. znan. umjet. 503. Matematickeˇ znanosti 16 (2009), str. 41–54 Mirko Radic´ and Zoran Kaliman About one relation concerning two circles Now from Figure 2 we see that t2 − r2 |A A |2 =(t +t )2 +(t +t )2 − 2(t +t )(t +t ) 2 , (6) 1 3 1 2 2 3 1 2 2 3 2 + 2 t2 r since 2 r 2 1 − 2 2 1 − tan β2 t2 t − r cos2β = = = 2 . (7) 2 1 + tan2 2 2 t2 + r2 β 1 + r 2 t2 The tangent length t2 =(t2)1 is given by (3) and tangent length t3 can be written as √ t (R2 − d2)+ D t = 2 2 (8a) 3 2 + 2 r t2 where = 2( 2 − 2)2 +( 2 + 2) 2 2 − 2 2 − ( 2 + 2 − 2)2 . D2 t2 R d r t2 4R d r t2 R d r (8b) Using computer, it can be found that 4 2 2 2 2 2 2 2 4 k(t1 +t3) −|A1A3| = 0 ⇐⇒ d k − 2d R k − 4r R + k R = 0. (9) 4 2 − 2 2 2 − 2 2 + 2 4 = = 2rR But d k 2d R k 4r R k R 0ifk R2−d2 . As will be seen in the following sections, analogously holds for the cases when one circle is not inside of the other. Only both circles have to be in the same plane. 2. The case when one circle is outside of the other The following theorem will be proved. THEOREM 1. Let C1 and C2 be any given two circles in the same plane such that C1 is outside of C2 .Letr= radius of C1 ,R= radius of C2 ,d= |IO|,whereIis the center of the C1 and O is the center of C2 Let A1A2A3 be any given triangle such that C2 is its circumcircle and that lines A1A2 and A2A3 be tangent lines to C1 .Their tangent points let be denoted by T1 and T2 respectively. Then 2rR |A A | =(t +t ) , (10) 1 3 1 3 d2 − R2 where t1 = |A1T1|,t3 = |A3T2|. (See Figure 3) Proof. First we prove how t2 = |A2T1| = |A1T2| and t3 can be calculated if t1 is given. For this purpose we prove the following lemma. LEMMA 1. If t1 is given then t2 can be calculated using expression √ t (d2 − R2)+ D t = 1 1 (11a) 2 2 + 2 r t1 where = 2( 2 − 2)2 +( 2 + 2) 2 2 − 2 2 − ( 2 + 2 − 2)2 . D1 t1 d R r t1 4d R r t1 d R r (11b) 43 Rad Hrvat. akad. znan. umjet. 503. Matematickeˇ znanosti 16 (2009), str. 41–54 Mirko Radic´ and Zoran Kaliman About one relation concerning two circles Figure 3 Proof. From triangles A1IT1 and A2IT2 ,whereA1(x1,y1), A2(x2,y2), I(d,0),we see that 2 =( − )2 + 2 − 2, 2 =( − )2 + 2 − 2 t1 x1 d y1 r t2 x2 d y2 r from which follows R2 + d2 − r2 −t2 R2 + d2 − r2 −t2 x = 1 , x = 2 . (12) 1 2d 2 2d Now, we have 2 2 2 2 2 (t1 −t2) = |A1A2| =(x1 − x2) +(y1 − y2) = 2R − 2x1x2 − 2y1y2 2 2 or 2R − 2x1x2 − (t1 −t2) = 2y1y2. The relation 2 2 2 2 2R − 2x1x2 − (t1 −t2) =(2y1y2) 2 + 2 = 2 + 2 = 2 using x1 y1 x2 y2 R and (12), can be written as ( − )2 ( 2 + 2) 2 − ( 2 − 2)+ 2 2 − 2 2 +( 2 + 2 − 2)2 = . t1 t2 r t1 t2 2t1t2 d R r t1 4R d R d r 0 We shall show that for every t1 which can be drawn from C2 to C1 we can find t2 as a solution of the equation ( 2 + 2) 2 − ( 2 − 2)+ 2 2 − 2 2 +( 2 + 2 − 2)2 = . r t1 t2 2t1t2 d R r t1 4R d R d r 0 (13) 2 2 For this purpose we shall in (13) replace t2 by t1 . The obtained equation can be written as 2 2 2 2 t1 − d +(R − r) t1 − d +(R + r) = 0. (14) Thus, if t1 is a solution of the above equation, then t2 = t1 . In this case triangle A1A2A3 is degenerate. (See Figure 4.) Also, it can be found that solutions of the equation (13) for t2 are given by (11) Thus Lemma 1 is proved. In this connection let us remark that, if t2 is given such that t2 = t1 ,thent3 is one of the solutions of the equation ( 2 + 2) 2 − ( 2 − 2)+ 2 2 − 2 2 +( 2 + 2 − 2)2 = . r t2 t3 2t2t3 d R r t2 4R d R d r 0 (15) 44 Rad Hrvat. akad. znan. umjet. 503. Matematickeˇ znanosti 16 (2009), str. 41–54 Mirko Radic´ and Zoran Kaliman About one relation concerning two circles Figure 4 The other solution is t1 = t2 . See, for example, Figure 4, where t3 = |A3T3|. As will be shown, for degenerate triangle A1A2A3 also holds 2rR |A A | =(t +t ) . 1 3 1 3 d2 − R2 2 2 In the case when t3 = 0,then 2Rr = d −R . Namely, in this case the relation is Euler’s relation for triangle where excircle instead of incircle is under consideration. (See, for example, [2, Corollary 2.1.3].) Now we prove Theorem 1. First from Figure 3 we see that t2 − r2 |A A |2 =(t −t )2 +(t −t )2 − 2(t −t )(t −t ) 2 , (16) 1 3 1 2 2 3 1 2 2 3 2 + 2 t2 r since 2 r 2 1 − 2 2 1 − tan β2 t2 t − r cos2β = = = 2 . (17) 2 1 + tan2 2 2 t2 + r2 β 1 + r 2 t2 The tangent length t2 =(t2)1 is given by (11) and tangent length t3 can be written as √ t (d2 − R2)+ D t = 2 2 , (18a) 3 2 + 2 r t2 where = 2( 2 − 2)2 +( 2 + 2) 2 2 − 2 2 − ( 2 + 2 − 2)2 . D2 t2 d R r t2 4R d r t2 R d r (18b) Using computer, it is not difficult to find that 4 2 2 2 2 2 2 2 4 k(t1 +t3) −|A1A3| = 0 ⇐⇒ d k − 2d R k − 4r R + k R = 0. (19) 4 2 − 2 2 2 − 2 2 + 2 4 = = 2rR But d k 2d R k 4r R k R 0ifk d2−R2 . This proves Theorem 1 Here are some illustrations. 45 Rad Hrvat. akad. znan. umjet. 503. Matematickeˇ znanosti 16 (2009), str. 41–54 Mirko Radic´ and Zoran Kaliman About one relation concerning two circles Figure 5 EXAMPLE 2.1. Let R = 3.8, r = 2.75, d = 7.7, t1 = 9.1. (See Figure 5.) Then = 2( 2 − 2)2 +( 2 + 2) 2 2 − 2 2 − ( 2 + 2 − 2)2 = . , D1 t1 d R r t1 4d R r t1 d R r 154 28498977 √ t (d2 − R2) ± D (t ) = 1 1 → (t ) = 6.223353231, (t ) = 2.808929822, (20) 2 1,2 2 + 2 2 1 2 2 r t1 (D ) =(t )2(d2 − R2)2 +(r2 +(t )2) 4R2d2 − r2(t )2 − (R2 + d2 − r2)2 , 2 1 2 1 2 1 2 1 ( ) =( )2( 2 − 2)2 +( 2 +( )2) 2 2 − 2( )2 − ( 2 + 2 − 2)2 , D2 2 t2 2 d R r t2 2 4R d r t2 2 R d r (D ) = 142.145499, (D ) = 14.63807559, 2 1 2 2 (t ) (d2 − R2) ± (D ) ((t ) ) = 2 1 2 1 3 1 1,2 r2 +(t )2 2 1 (t ) (d2 − R2) ± (D ) ((t ) ) = 2 2 2 2 3 2 1,2 2 +( )2 r t2 2 (( ) ) = .
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