Intro to Circle Geometry By Raymond Cheong


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1 Into to Cicle Geomety By Rymond Cheong Mny poblems involving cicles cn be solved by constucting ight tingles then using the Pythgoen Theoem. The min chllenge is identifying whee to constuct the ight tingle. As hint, sides of the tingle e often: A line segment connecting the cente of two cicles tht e tngent. This line psses though the tngent point nd its length is the sum o the diffeence of the dii, depending on whethe the cicles e extenlly o intenlly tngent, espectively. A dius to tngent point. The dius nd the tngent fom ight ngle. A line pllel o pependicul to existing lines, especilly to sides of sques o ectngles. This helps to genete ight ngles. Afte constucting the tingle, lbel the sides in tems of, sy, the unknown dius. Then use the Pythgoen Theoem to solve fo. Thee e usully few poblems evey ye in the Mylnd Mth Legue tht cn be solved in this wy. Conside the following exmple, which ws slightly modified fom the MML (poblem 64): Poblem: Two extenlly tngent cicles e conguent. A line segment is dwn fom the cente of one cicle, tngent to the othe. If the length of this segment is, wht is the dius of one of the cicles? Solution: Cete the ight tingle shown. One leg is the tngent with length. One leg is dius to the tngent, nd denote its length. The hypotenuse connects the centes of the cicle nd its length is. By the Pythgoen Theoem, then, + =, giving =.
2 Poblems ) Two pependicul lines, intesecting t the cente of cicle of dius, divide the cicle into fou pts. A smlle cicle is inscibed in one of those pts s shown. Wht is the dius of the smlle cicle? ) A point on cicle inscibed in sque is nd units fom the two closest sides of the sque. Find the e of the sque. 3) A cicle of dius 9 is extenlly tngent to cicle of dius 6. Find the length of thei common extenl tngent. 4) In the figue shown, chod of length 6 is pependicully bisected by line segment of length. Find the dius of the cicle ) In the figue, cicle of dius is inscibed in sque. A smlle cicle is tngent to two sides of the sque nd the fist cicle. Detemine the dius of the smlle cicle. 6) Thee conguent cicles e plced inside semicicle such they e tngent to the bse of the semicicle nd to ech othe s shown. If the semicicle hs dius 4, find the dius of one of the cicles. 7) As shown in the figue, cicle of dius is tngent to semicicle nd the cente of its bse. A smlle cicle is tngent to the cicle, semicicle, nd its bse. Detemine the dius of the smlle cicle. 8) In the figue shown, AB is dimete of cicle, nd AC, CD, nd BD e tngent to the cicle t points A, E, nd B, espectively. If CEDE = 3, detemine the e of the cicle. C A E D B
3 Solutions ) [003 TAMU DE #] The key insight in this poblem is to use the line connecting the centes of the two cicles s the hypotenuse of ight tingle. Let be the dius of the smlle cicle. Conside the dius of the lge cicle tht psses though the cente of the smlle cicle. This dius hs length, nd is split into segments of nd . Theefoe, the ight tingle in question hs two legs nd hypotenuse , giving + = (). Solve to get = (the othe oot is negtive). ) [00 HMMT Genel #8]  Use the dius to the point in question s the hypotenuse of ight tingle, nd use legs pllel to the sides of the sques. If the dius is, then the legs e  nd . Thus, () + () =, nd simplify to ()(5) = 0. If = then the sque is nd point on the cicle could not possibly be distnce fom n edge. So, = 5. The e is () = 0 = ) [Clssic] We will use the line segment connecting the centes of the cicles s the hypotenuse of ight tingle. 9 To constuct this tingle, begin by dwing dii to the points of tngency. The dii meet the extenl 9 7 tngent t ight ngles. Next, dw line pllel to 6 9 the extenl tngent fom the cente of the smlle cicle to the dius of the lge cicle. This cetes ectngle nd ight tingle. The ight tingle hs hypotenuse of = 5 nd leg of 6 9 = 7. So the othe leg is 4 ( tingle), which is equl to the length of the extenl tngent. 4) [Modified fom 00 Tennessee Mth Contest Femt I #0] Constuct the ight tingle s shown, with dius s the hypotenuse nd hlf of the chod, 3, s leg. The othe leg is the sme s dius minus the pependicul bisecto, o . So, 3 + () =. Solve to get = 3/
4 5) [Modified fom 004 UNCW #] Use the line segment connecting the centes of the cicles s the hypotenuse of ight tingle. Dw the legs pllel to the sides of the sques. If we denote the dius of the smlle cicle s, then the hypotenuse hs length +. Since the cente of the smlle cicle is distnce fom the two closest sides, the legs hve length . By the Pythgoen Theoem, (+) = () + (). Renge to 6 + = 0, nd use the qudtic fomul to get = 3. The positive oot is gete thn, so = ) [005 Indin Stte Mth Contest, Geomety/Integted Mth II, #] Use the line segment connecting the cente of the semicicle nd the ight hnd cicle s the hypotenuse of ight tingle. One leg is the dius of the middle cicle. The othe leg connects the centes of the middle nd ight hnd cicles. If we denote the unknown dius to be, then + () = (4). Expnd nd enge to 4( + 4) = 0. The qudtic fomul gives =  5. The negtive oot is negtive, so the nswe is ) [006 TAMU DE #] 4 This poblem cn be solved by constucting two ight tingles. The hypotenuse of the fist tingle connects the centes of the semicicle nd the smlle cicle (lowe ight in digm). If we denote the dius of the smlle cicle to be, then the hypotenuse hs length 4. One of the legs of this tingle is dius of the smlle cicle, denoted, nd the othe leg is denoted x. Thus, + x = (4).  x + 4 x The hypotenuse second ight tingle connects the centes of the two cicles, fo length of +. The legs e  nd x, giving () + x = (+). Solve fo x in both equtions to eliminte it: x = (4) = (+) (). Expnd nd solve to give =.
5 8) [Modified fom NSML Mth Contest smple, #3] Dw the pependicul fom C to BD, foming ectngle nd ight tingle. Denote the dius of the cicle to be, CE = nd DE = b. Since tngents to common point e equl, AC = CE = nd BD = DE = b. Then, the ight tingle hs legs nd b nd hypotenuse +b, giving () + (b) = (+b). Expnd nd note tht the nd b tems cncel out, nd the est simplifies to b = = 3. So, the e is = 3. C A E b D b B
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