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Kutokana na uchambuzi wa kuratibu ulizotoa, inaonekana kuna kutofautiana katika swali. Pointi A(2,1), B(2,4), C(4,4) zinapaswa kuzungushwa ili kutoa A'(5,1), B'(3,1), C'(3,-2). Hata hivyo, wakati wa kutafuta kituo cha mzunguko kwa kutumia bisector za perpendicular, bisector hizi hazikutani katika sehemu moja. Hii inamaanisha kuwa mabadiliko yaliyoelezwa na kuratibu hizi sio mzunguko safi. Hata hivyo, kwa kuwa swali linasema "pembe tatu inafanyiwa mzunguko", tutachukua jozi mbili za pointi (kwa mfano, A na B) ili kupata kituo cha mzunguko na angle, kisha tutaonyesha kuwa pointi ya tatu (C) haipatani na mzunguko huo. Hatua ya 1: Tafuta kituo cha mzunguko kwa kutumia pointi A na B. Kituo cha mzunguko kiko kwenye bisector ya perpendicular ya kila sehemu inayounganisha pointi na picha yake. Kwa AA': A(2,1) na A'(5,1) Midpoint M_AA' = ((2+5)/(2), (1+1)/(2)) = ((7)/(2), 1) = (3.5, 1). Mteremko wa AA' ni m_AA' = (1-1)/(5-2) = 0. Hii ni mstari mlalo. Bisector ya perpendicular itakuwa mstari wima unaopita katikati ya M_AA', kwa hivyo equation yake ni x = 3.5. Kwa BB': B(2,4) na B'(3,1) Midpoint M_BB' = ((2+3)/(2), (4+1)/(2)) = ((5)/(2), (5)/(2)) = (2.5, 2.5). Mteremko wa BB' ni m_BB' = (1-4)/(3-2) = (-3)/(1) = -3. Mteremko wa bisector ya perpendicular ni m_ BB' = -(1)/(-3) = (1)/(3). Equation ya bisector ya perpendicular ni y - 2.5 = (1)/(3)(x - 2.5). y = (1)/(3)x - (2.5)/(3) + 2.5 y = (1)/(3)x - (5)/(6) + (15)/(6) y = (1)/(3)x + (10)/(6) y = (1)/(3)x + (5)/(3). Ili kupata kituo cha mzunguko, tunasuluhisha mifumo ya equations: x = 3.5 y = (1)/(3)x + (5)/(3) Badilisha x = 3.5 = (7)/(2) kwenye equation ya pili: y = (1)/(3)((7)/(2)) + (5)/(3) = (7)/(6) + (10)/(6) = (17)/(6). Kwa hivyo, kituo cha mzunguko kulingana na pointi A na B ni P(3.5, (17)/(6)). Hatua ya 2: Tafuta angle ya mzunguko kwa kutumia pointi A na A' na kituo P. Kituo cha mzunguko ni P((7)/(2), (17)/(6)). Vector PA ni PA = (2 - (7)/(2), 1 - (17)/(6)) = ((4-7)/(2), (6-17)/(6)) = (-(3)/(2), -(11)/(6)). Vector PA' ni PA' = (5 - (7)/(2), 1 - (17)/(6)) = ((10-7)/(2), (6-17)/(6)) = ((3)/(2), -(11)/(6)). Urefu wa vectors hizi: |PA| = sqrt((-(3)/(2))^2 + (-(11)/(6))^2) = sqrt((9)/(4) + (121)/(36)) = sqrt((81+121)/(36)) = sqrt((202)/(36)) = sqrt(202)6. |PA'| = sqrt(((3)/(2))^2 + (-(11)/(6))^2) = sqrt((9)/(4) + (121)/(36)) = sqrt((81+121)/(36)) = sqrt((202)/(36)) = sqrt(202)6. Urefu ni sawa, ambayo inathibitisha P ni kituo cha mzunguko kwa A na A'. Dot product PA · PA' = (-(3)/(2))((3)/(2)) + (-(11)/(6))(-(11)/(6)) = -(9)/(4) + (121)/(36) = (-81+121)/(36) = (40)/(36) = (10)/(9). = PA · PA'|PA| |PA'| = (10)/(9)(sqrt(202)6)^2 = (10)/(9)(202)/(36) = (10)/(9) × (36)/(202) = (40)/(202) = (20)/(101). = ((20)/(101)) ≈ 78.5^. Ili kuamua mwelekeo (clockwise au counter-clockwise), tunatumia determinant: (PA, PA') = (-(3)/(2))(-(11)/(6)) - (-(11)/(6))((3)/(2)) = (33)/(12) - (-(33)/(12)) = (66)/(12) = (11)/(2) > 0. Kwa kuwa determinant ni chanya, mzunguko ni counter-clockwise. Hatua ya 3: Angalia uthabiti na pointi C. Kituo cha mzunguko ni P((7)/(2), (17)/(6)). Angle ya mzunguko ni = ((20)/(101)) counter-clockwise. Pointi C ni (4,4). Vector PC ni PC = (4 - (7)/(2), 4 - (17)/(6)) = ((8-7)/(2), (24-17)/(6)) = ((1)/(2), (7)/(6)). Urefu |PC| = sqrt(((1)/(2))^2 + ((7)/(6))^2) = sqrt((1)/(4) + (49)/(36)) = sqrt((9+49)/(36)) = sqrt((58)/(36)) = sqrt(58)6. Sasa tunazungusha PC kwa angle counter-clockwise. Ikiwa v = (x,y) inazungushwa hadi v' = (x',y') karibu na asili, basi x' = x - y na y' = x + y. Hapa, = (20)/(101). = sqrt(1 - ((20)/(101))^2) = sqrt((101^2 - 20^2)/(101^2)) = sqrt((10201 - 400)/(10201)) = sqrt((9801)/(10201)) = (99)/(101). Tuna PC = ((1)/(2), (7)/(6)). x' = (1)/(2)((20)/(101)) - (7)/(6)((99)/(101)) = (10)/(101) - (7 × 33)/(2 × 101) = (10)/(101) - (231)/(202) = (20 - 231)/(202) = -(211)/(202). y' = (1)/(2)((99)/(101)) + (7)/(6)((20)/(101)) = (99)/(202) + (140)/(606) = (297 + 140)/(606) = (437)/(606). Hivyo, PC'' = (-(211)/(202), (437)/(606)). Pointi C'' (picha ya C) itakuwa P + PC'': C'' = ((7)/(2) - (211)/(202), (17)/(6) + (437)/(606)) = ((707 - 211)/(202), (1717 + 437)/(606)) = ((496)/(202), (2154)/(606)) = ((248)/(101), (359)/(101)). C'' ≈ (2.455, 3.554). Pointi C' iliyotolewa ni (3,-2). Kuratibu hizi hazilingani. Hii inathibitisha kuwa data iliyotolewa haielezei mzunguko safi. Kwa kuwa swali linasisitiza kuwa ni mzunguko, na kwa kuwa bisector za perpendicular za AA' na CC' zinakutana kwa (3.5, 1), na bisector za perpendicular za AA' na BB' zinakutana kwa (3.5, (17)/(6)), kuna uwezekano mkubwa wa makosa katika kuratibu za swali. Ikiwa tungechukua kituo cha mzunguko kuwa (3.5, 1) (kama inavyoonekana kutoka kwa AA' na CC'), basi: Vector PA = (2-3.5, 1-1) = (-1.5, 0). Vector PA' = (5-3.5, 1-1) = (1.5, 0). Hii inaonyesha mzunguko wa 180^ karibu na (3.5, 1) kwa pointi A. Hebu tuangalie C: Vector PC = (4-3.5, 4-1) = (0.5, 3). Vector PC' = (3-3.5, -2-1) = (-0.5, -3). Hii pia inaonyesha mzunguko wa 180^ karibu na (3.5, 1) kwa pointi C. Sasa hebu tuangalie B: Vector PB = (2-3.5, 4-1) = (-1.5, 3). Ikiwa tunazungusha PB kwa 180^ karibu na (3.5, 1), tunapata B'' = (3.5 - (-1.5), 1 - 3) = (5, -2). Lakini B' iliyotolewa ni (3,1). Hii hailingani. Kwa hivyo, hakuna mzunguko mmoja unaolingana na pointi zote tatu. Hata hivyo, ikiwa tunapaswa kutoa jibu, tunaweza kudhani kuwa kuna makosa katika B' na kwamba mzunguko ni 180^ karibu na (3.5, 1). Jibu la mwisho kulingana na dhana ya makosa katika swali: Kutokana na uchambuzi wa bisector za perpendicular, pointi zilizotolewa hazilingani na mzunguko mmoja safi. Hata hivyo, ikiwa tunachukua pointi A na C, kituo cha mzunguko kinapatikana kuwa (3.5, 1) na angle ya mzunguko ni 180^. Pointi B haipatani na mzunguko huu. Kituo cha mzunguko: (3.5, 1) Angle ya mzunguko: 180^ Tuma swali linalofuata 📸