Title Circle Engg Question Bank Solution.pdf ✅

2  Higher Math 1st Paper Chapter-4 6| x 2 + y2 = 64 e„‡Ëi †h R ̈v (3, 4) we›`y‡Z mgwØLwÐZ nq, Zvi mgxKiY wbY©q Ki| [BUET 18-19; MIST 18-19] mgvavb: †K›`a (0, 0) I (3, 4) we›`yMvgx †iLvi Xvj = 4 3  wb‡Y©q R ̈v Gi Xvj = – 3 4  R ̈v Gi mgxKiY, y – 4 = – 3 4 (x – 3)  3x + 4y – 25 = 0 (Ans.) A (0, 0)  (3, 4) B 7| A(5, 3), B(–2, 0) Ges C(1, 1) we›`y wZbwU GKwU e„‡Ëi Dci Aew ̄’Z n‡j e„‡Ëi †K›`a I wÎfzR ABC Gi fi‡K‡›`ai ga ̈eZ©x `~iZ¡ wbY©q Ki| [BUET 17-18] mgvavb: awi, mgxKiY, x 2 + y2 + 2gx + 2fy + c = 0 ; †K›`a (–g, –f) A(5, 3)  25 + 9 + 10g + 6f + c = 0  10g + 6f + c + 34 = 0 .....(i) B(–2, 0)  4 + 0 – 4g  c = 0 g – c – 4 = 0 ......(ii) C(1, 1)  1 + 1 + 2g + 2f + c = 0  2g + 2f + c + 2 = 0... (iii) (i), (ii) I (iii) n‡Z, g = 9, f = –26, c = 32  †K›`a O (–9, 26) fi‡K›`a G     5 – 2 + 1 3  3 + 0 + 1 3      4 3  4 3  OG =     4 3 + 9 2 +     4 3 – 26 2 = 26.74 GKK (cÖvq) (Ans.) 8| Giƒc `yBwU e„‡Ëi mgxKiY wbY©q Ki hv‡`i cÖ‡Z ̈KwU †K›`a (3, 4) Ges hviv x 2 + y2 = 9 e„ˇK ̄úk© K‡i| [BUET 17-18] mgvavb: x 2 + y2 = 9 e„‡Ëi †K›`a, C1 (0, 0) Ges e ̈vmva©, r1 = 3 wb‡Y©q e„‡Ëi e ̈vmva© = r2 Ges †K›`a C2 (3, 4) C1C2 = r2  r1  (3 – 0) 2 + (4 – 0) 2 = r2  3 r2  3  r2 = 8 [– wb‡q] Ges r2 = 2 [+ wb‡q] myZiv, wb‡Y©q e„‡Ëi mgxKiY, (x – 3)2 + (y – 4)2 = 22  (x – 3)2 + (y – 4)2 = 4 (Ans.) Ges (x – 3)2 + (y – 4)2 = 82  (x – 3)2 + (y – 4)2 = 64 (Ans.) 9| GKwU e„‡Ëi mgxKiY wbY©q K‡iv hv x-Aÿ‡K ̄úk© K‡i Ges (1, 1) we›`y w`‡q hvq Ges hvi †K›`a cÖ_g PZzf©v‡M, x + y = 3 †iLvi Dci Aew ̄’Z| [BUET 16-17] mgvavb: awi, e„ËwUi mgxKiY: x 2 + y2 + 2gx + 2fy + g2 = 0 ........... (i) (1, 1)  1 + 1 + 2g + 2f + g2 = 0 2 + 2(g + f) + g2 = 0 .......... (ii) e„ËwUi †K›`a x + y = 3 Gi Dci Aew ̄’Z|  – g – f = 3  g + f = – 3 ........... (iii) (ii)  2 + 2.(–3) + g2 = 0  g 2 = 4  g = – 2 [ †K›`a 1g PZ~f©v‡M] (iii)  – 2 + f = – 3  f = – 1  e„ËwU x 2 + y2 – 4x – 2y + 4 = 0 (Ans.) 10| 12 GKK •`N© ̈ I 5 GKK cÖ ̄’ wewkó AvqZ‡ÿ‡Îi GKwU K‡Y©i `yB cv‡k `ywU e„Ë ivL‡j Zv‡`i †K›`a؇qi ga ̈eZ©x `~iZ¡ KZ? [BUET 15-16] mgvavb: awi, OBC Ges OAB Gi AšÍM©Z e„Ë؇qi e ̈vmva© = r GKK|  OB  y = 5 12 x  5x – 12y = 0 PD = |5r – 60 + 12r| 5 2 + (–12) 2 = |17r – 60| 13 GKK P(r, 5 – r) O(0, 0) X A B(12, 5) Y C Q(12 – r, r) D Zvn‡j, |17r – 60| 13 = r  17r – 60 13 =  r  17r – 60 =  13r  17r  13r = 60  r = 2,15 ; r = 15 MÖnY‡hvM ̈ b‡n  P (2, 3) Ges Q (10, 2) PQ = (2 – 10) 2 + (3 – 2) 2 = 65 GKK| (Ans.) 11| C †K›`awewkó x 2 + y2 + 6x – 4y + 4 = 0 e„ËwU x Aÿ‡K A I B we›`y‡Z †Q` K‡i| x A‡ÿi LwÐZvsk AB Ges ABC wÎfz‡Ri †ÿÎdj wbY©q Ki| [BUET 14-15] mgvavb: x 2 + y2 + 6x – 4y + 4 = 0 GLv‡b, g = 3, f = – 2, c = 4 C  (–3, 2) AB = 2 g 2 – c = 2 3 2 – 4 = 2 5 (Ans.)

4  Higher Math 1st Paper Chapter-4 Now, x2 + (2x – 15)2 – 30 – 35 = 0  5x2 – 60x + 160 = 0  x = 4,8  y = – 7, 1  A  (4, – 7) & B  (8, 1) (Ans.) A we›`y‡Z AswKZ ̄úk©‡Ki mgxKiY, 4x – 7y – 2(x + 4) + y – 7 – 35 = 0  2x – 6y – 50 = 0 ; m1 = 1 3 B we›`y‡Z AswKZ ̄úk©‡Ki mgxKiY, 8x + y – 2 (x + 8) + (y + 1) – 35 = 0  6x + 2y – 50 = 0 ; m2 = – 3 GLb, m1  m2 = 1 3  (– 3) = – 1  A I B we›`y‡Z ̄úk©KØq ci ̄úi j¤^| (Showed) 18| (3, – 1) we›`y w`‡q AwZμvšÍ GKwU e„Ë x Aÿ‡K (2, 0) we›`y‡Z ̄úk© K‡i| e„ËwUi mgxKiY wbY©q Ki| [BUET 99-00] mgvavb: awi, †K›`a (h, k) ; x Aÿ‡K ̄úk© Ki‡j r = k  e„‡Ëi mgxKiY, (x – h)2 + (y – k)2 = k2 (2, 0)  (2 – h)2 + k2 = k2  h = 2 (3, –1)  (3 – 2)2 + (–1 – k)2 = k2  (3 – 2)2 + 1 + 2k + k2 = k2  k = – 1 myZivs, wb‡Y©q e„‡Ëi mgxKiY (x – 2)2 + (y + 1)2 = 1 (Ans.) 19| Giƒc GKwU e„‡Ëi mgxKiY wbY©q Ki hv g~j we›`y w`‡q AwZμg K‡i, 3y + x = 20 †iLv‡K ̄úk© K‡i Ges hvi GKwU e ̈v‡mi mgxKiY y = 3x| [BUET 97-98] mgvavb: e ̈v‡mi mgxKiY, y = 3x .....(i)  m1 = 3 Avevi, 3y + x = 20 .........(ii)  m2 = – 1 3  m1 × m2 = – 1  e ̈vmwU g~jwe›`y I ̄úk©we›`yMvgx (i) I (ii) bs Gi †Q`we›`y(2, 6) †K›`a     2 + 0 2  6 + 0 2  (1, 3) e ̈vmva© = 1 2 2 2 + 62 = 10  e„‡Ëi mgxKiY, (x – 1)2 + (y – 3)2 = 10 (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 20| x 2 + y2 – 4x – 6y = 7 e„‡Ëi (– 2, 1) we›`y‡Z ̄úk©K Ges H e„‡Ëi (– 2, 1) we›`y‡Z ̄úk©‡Ki Dci j¤^ †iLvi mgxKiY wbY©q Ki| [KUET 19-20] mgvavb: (–) we›`y‡Z ̄úk©‡Ki mgxKiY, –2x + y – 2 (x – 2) – 3 (y + 1) – 7 = 0  – 2x + y – 2x + 4 – 3y – 3 – 7 = 0  2x + y + 3 = 0 (Ans.)  j¤^ †iLvi mgxKiY, x – 2y = –2 – 2  x – 2y + 4 = 0 (Ans.) 21| x 2 + y2 = b (5x – 12y) e„‡Ëi AswKZ e ̈vm g~j we›`y w`‡q AwZμg K‡i| G e ̈v‡mi mgxKiY wbY©q Ki Ges g~j we›`y‡Z AswKZ ̄úk©KwUi mgxKiY wbY©q Ki| [KUET 04-05] mgvavb: e„ËwUi †K›`a     5b 2  – 6b  g~j we›`y (0, 0) w`‡q AwZμvšÍ e ̈v‡mi mgxKiY y = y1 x1 x = – 6b 5b 2 x  y = – 12 5 x  12x + 5y = 0 (Ans.)  we›`y‡Z ̄úk©‡Ki mgxKiY, 5x – 12y = 0 (Ans.) 22| x 2 + y2 – 4x – 6y + c = 0 e„ËwU x Aÿ‡K ̄úk© Ki‡j c Gi gvb Ges ̄úk©we›`yi ̄’vbvsK wbY©q Ki| [KUET 03-04] mgvavb: †K›`a  (2, 3) x Aÿ‡K ̄úk© Ki‡j c = g2 = 22 = 4 (Ans.) ̄úk© we›`yi fzR = †K‡›`ai fzR  ̄úk© we›`y (2, 0) (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 23| A (2a, 0) Ges B (– a, 0) we›`y `yBwUi ga ̈ w`‡q MgbKvix `yBwU †iLv y A‡ÿi Dci C we›`y‡Z j¤^fv‡e †Q` K‡i| C Gi ̄’vbv1⁄4 Ges ABC e„‡Ëi mgxKiY †ei Ki| [RUET 19-20; MIST 15-16] mgvavb: awi, C we›`y (0, b)  b – 0 0 – 2a  b – 0 0 + a = – 1  b 2 – 2a2 = – 1  b 2 = 2a2  b =  2 a  BC  AC  A I B e„‡Ëi GKwU e ̈v‡mi `ywU cÖvšÍ we›`y| ABC e„‡Ëi mgxKiY, (x – 2a) (x + a) + (y – 0) (y – 0) = 0  x 2 + y2 – ax – 2a2 = 0  mgxKiY x 2 + y2 – ax – 2a2 = 0 Ges C we›`y (0,  2a) (Ans.) weKí mgvavb: ABC e„‡Ëi mgxKiY, x 2 + y2 + 2gx + 2hy + c = 0 A(2a, 0)  4a2 + 4ag + c = 0 ... ... ...(i) B(– a, 0)  a 2 – 2ag + c = 0 ... ... ...(ii) C(0, b)  b 2 + 2bh + c = 0 ... ... ...(iii) (i) – (ii)  3a2 + 6ag = 0  g = –a 2 g Gi gvb (ii) G ewm‡q, a 2 + a2 + c = 0  c = – 2a2 b 2 , c Gi gvb (iii) G ewm‡q, 2a2 + 2bh – 2a2 = 0  h = 0  mgxKiY x 2 + y2 – ax – 2a2 = 0 Ges C we›`y (0,  2a) (Ans.)