Tiêu đề Circle Engineering Practice Sheet Solution.pdf ✅

e„Ë  Engineering Practice Sheet Solution 1 04 e„Ë The Circle WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. x 2 + y2 = 64 e„‡Ëi †h R ̈v (3, – 3) we›`y‡Z mgwØLwÐZ nq, Zvi mgxKiY wbY©q Ki| [BUET 23-24] mgvavb: OC †iLvi Xvj = 0 + 3 0 – 3 = – 1  AB †iLvi Xvj = 1 †iLvi mgxKiY, y + 3 = 1(x – 3)  y + 3 = x – 3  x – y – 6 = 0 (Ans.) O(0, 0) A C(3, – 3) B 2. 12 dzU e ̈vmv‡a©i GKwU Aa©e„ËvKvi myo‡1⁄2i wfZi w`‡q `yB †j‡bi GKwU iv ̄Ív wM‡q‡Q, hvi cÖwZwU 10 dzU cÖk ̄Í| iv ̄Ívi wKbvivq myo‡1⁄2i D”PZv KZ? [BUET 22-23] mgvavb: OA2 = AB2 + OB2  AB = 122 – 102 = 44 = 2 11 dzU (Ans.) A   O 10 B 12 weKí mgvavb: (x – 0)2 + (y – 0)2 = 122  x 2 + y2 = 144 (10, h) Øviv wm× K‡i, 102 + h2 = 144  h = 44 = 2 11 dzU (Ans.)  (10, h)    (– 12, 0) (0, 0) (12, 0) 3. wZbwU e„Ë ci ̄úi‡K ̄úk© K‡i Ges G‡`i GKwU mvaviY ̄úk©K Av‡Q| eo e„ËwUi e ̈vmva© = 4 GKK| gvSvwi e„ËwUi e ̈vmva© = 2 GKK n‡j, †QvU e„ËwUi e ̈vmva© †ei Ki| [BUET 21-22] mgvavb: A M P N D R Q B c r1 r2 r3 r1 r2 GLv‡b, r1 = 4, r2 = 2, r3 = ? MN2 = BD2 = AB2 – AD2 = (r1 + r2) 2 – (r1 – r2) 2 = 4r1r2  MN = 2 r1r2 Abyiƒcfv‡e, PM = 2 r1r3 Ges PN = 2 r2r3 MN = PM + PN  2 r1r2 = 2 r1r3 + 2 r2r3  2 4 = 4r3 + 2r3 r3 = 0.686 GKK| (Ans.) 4. (9, 8) †K›`awewkó e„Ë x 2 + y2 – 2x – 4y – 20 = 0 e„ˇK ewnt ̄’fv‡e ̄úk© Ki‡j e„‡Ëi mgxKiY wbY©q Ki| [BUET 20-21] mgvavb:  C1(1, 2)  C2(9, 8) r1 = 5 ewnt ̄’fv‡e ̄úk© Ki‡j, r1 + r2 = C1C2 = (9 – 1) 2 + (8 – 2) 2 = 10 r2 = 10 – (– 1) 2 + (– 2) 2 + 20 = 5  (9, 8) †K›`a I e ̈vmva© 5 wewkó e„‡Ëi mgxKiY, (x – 9)2 + (y – 8)2 = 52 (Ans.) 5. GKwU e„Ë (– 1, – 1) Ges (3, 2) we›`y w`‡q AwZμg K‡i Ges Gi †K›`a x 2 + y2 – 6x – 4y – 7 = 0 e„‡Ëi (1, – 2) we›`y‡Z ̄úk©‡Ki Dci Aew ̄’Z| e„ËwUi mgxKiY wbY©q Ki| [BUET 19-20; MIST 19-20] mgvavb: awi, e„‡Ëi †K›`a (, )  e ̈vmva© = ( + 1) 2 + ( + 1) 2 = ( – 3) 2 + ( – 2) 2   2 + 2 + 1 +  2 + 2 + 1 =  2 – 6 + 9 +  2 – 4 + 4  8 + 6 – 11 = 0 ...... (i) Avevi, (1, – 2) we›`ywU x 2 + y2 – 6x – 4y – 7 = 0 e„‡Ëi evgc‡ÿ ewm‡q cvB, 1 2 + (– 2)2 – 6  1 – 4(– 2) – 7 = 0  wb‡Y©q ̄úk©‡Ki mgxKiY, x – 2y – 3(x + 1) – 2(y – 2) – 7 = 0  x + 2y + 3 = 0   + 2 + 3 = 0 ........ (ii)
2  Higher Math 1st Paper Chapter-4 (i) I (ii) mgvavb K‡i cvB, (, )      4  – 7 2  e ̈vmva©, r = (4 + 1) 2 +    –  7 2 + 1 2 = 5 5 2 GKK  e„ËwUi mgxKiY, (x – 4)2 +     y + 7 2 2 = 125 4 (Ans.) 6. GKwU e„Ë (– 1, – 1) Ges (3, 2) we›`y w`‡q AwZμg K‡i Ges Gi †K›`a x 2 + y2 – 6x – 4y – 7 = 0 e„‡Ëi (1, – 2) we›`y‡Z ̄úk©‡Ki Dci Aew ̄’Z| e„ËwUi mgxKiY wbY©q Ki| [BUET 19-20; MIST 19-20] mgvavb: awi, e„‡Ëi †K›`a (, )  e ̈vmva© = ( + 1) 2 + ( + 1) 2 = ( – 3) 2 + ( – 2) 2   2 + 2 + 1 +  2 + 2 + 1 =  2 – 6 + 9 +  2 – 4 + 4  8 + 6 – 11 = 0 ...... (i) Avevi, (1, – 2) we›`ywU x 2 + y2 – 6x – 4y – 7 = 0 e„‡Ëi evgc‡ÿ ewm‡q cvB, 1 2 + (– 2)2 – 6  1 – 4(– 2) – 7 = 0  wb‡Y©q ̄úk©‡Ki mgxKiY, x – 2y – 3(x + 1) – 2(y – 2) – 7 = 0  x + 2y + 3 = 0   + 2 + 3 = 0 ........ (ii) (i) I (ii) mgvavb K‡i cvB, (, )      4  – 7 2  e ̈vmva©, r = (4 + 1) 2 +    –  7 2 + 1 2 = 5 5 2 GKK  e„ËwUi mgxKiY, (x – 4)2 +     y + 7 2 2 = 125 4 (Ans.) 7. 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) (3, 4) we›`yMvgx e ̈v‡mi Xvj, m1 = 4 3 R ̈vwUi Xvj = m2 m1 × m2 = – 1  m2 = – 3 4  R ̈v Gi mgxKiY, y – 4 = – 3 4 (x – 3)  4y – 16 = – 3x + 9  3x + 4y – 25 = 0 (Ans.) 8. 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) (3, 4) we›`yMvgx e ̈v‡mi Xvj, m1 = 4 3 R ̈vwUi Xvj = m2 m1 × m2 = – 1  m2 = – 3 4  R ̈v Gi mgxKiY, y – 4 = – 3 4 (x – 3)  4y – 16 = – 3x + 9  3x + 4y – 25 = 0 (Ans.) 9. 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 (cÖvq) (Ans.) 10. 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]  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.) 11. GKwU e„‡Ëi mgxKiY wbY©q Ki 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; BUTex 07-08] 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)
e„Ë  Engineering Practice Sheet Solution 3 (ii)  2 + 2(– 3) + g2 = 0  g 2 = 4  g = – 2 [ †K›`a 1g PZzf©v‡M] (iii)  – 2 + f = – 3  f = – 1  e„ËwUi mgxKiY, x 2 + y2 – 4x – 2y + 4 = 0 (Ans.) 12. 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: P(r, 5 – r) O(0, 0) X A B(12, 5) Y C Q(12 – r, r) D 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 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.) 13. 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 GKK (Ans.) A D B C Y X CD = C we›`yi †KvwU = 2 GKK  ABC Gi †ÿÎdj = 1 2  AB  CD = 1 2  2 5  2 = 2 5 eM©GKK (Ans.) weKí: A I B we›`yi †KvwU 0| GLb, x 2 + 6x + 4 = 0  x = – 6  2 5 2  x = – 3  5 ; C  (– 3, 2) GLb, ABC = 1 2       – 3 – 3 + 5 – 3 – 5 2 0 0 1 1 1 = 2 5 eM© GKK (Ans.) 14. GKwU e„‡Ëi mgxKiY wbY©q Ki hv y = 2 †iLv‡K (3, 2) we›`y‡Z ̄úk© Ges (1, 4) we›`y w`‡q hvq| [BUET 13-14] mgvavb: O C (h, k) (1, 4) (3, 2) Y 2 y = 2 X h = 3, k = r + 2, r = k – 2  (x – 3) 2 + (y – k)2 = (k – 2)2  2 2 + (4 – k)2 = (k – 2)2 [ (1, 4) we›`yMvgx]  4 + k2 – 8k + 16 = k2 – 4k + 4  k = 4, r = 2  wb‡Y©q mgxKiY, (x – 3)2 + (y – 4)2 = 4 (Ans.) 15. GKwU e„‡Ëi mgxKiY wbY©q Ki hv x-Aÿ‡K (4, 0) we›`y‡Z ̄úk© K‡i Ges y Aÿ n‡Z 6 GKK `xN© GKwU R ̈v LwÐZ K‡i| [BUET 11-12, 02-03] mgvavb: wP‡Î, CD  AB GLv‡b, BD = 6 2 = 3 OP = CD = 4  CB = CP = 4 2 + 32 = 5  †K‡›`ai ̄’vbv1⁄4 (4,  5) Ges e ̈vmva© = 5  e„‡Ëi mgxKiY, (x – 4)2 + (y  5)2 = 52 (Ans.) A D C B P O (4, 0) Y X
4  Higher Math 1st Paper Chapter-4 16. GKwU e„Ë y Aÿ‡K g~jwe›`y‡Z ̄úk© K‡i Ges (3, – 4) we›`y w`‡q AwZμg K‡i| e„ËwUi mgxKiY wbY©q Ki| [BUET 07-08; MIST 17-18] mgvavb: awi, e„‡Ëi mgxKiY, x 2 + y2 + 2gx + 2fy + c = 0 y Aÿ‡K g~jwe›`y‡Z ̄úk© K‡i ZvB f 2 = c = 0  mgxKiY, x 2 + y2 + 2gx = 0 (3, – 4)  9 + 16 + 2g × 3 = 0  g = – 25 6 e„ËwUi mgxKiY, x 2 + y2 + 2    –  25 6 x = 0  3x2 + 3y2 – 25x = 0 (Ans.) 17. (1, 2) †K›`a wewkó GKwU e„Ë x Aÿ‡K ̄úk© K‡i| Bnvi mgxKiY I y Aÿ‡K Bnv wK cwigvY Ask †Q` K‡i, Zv wbY©q Ki| [BUET 02-03] mgvavb: e„‡Ëi e ̈vmva© = †K‡›`ai †KvwU = 2 GKK  wb‡Y©q e„‡Ëi mgxKiY, (x – 1)2 + (y – 2)2 = 22  x 2 + y2 – 2x – 4y + 1 = 0  y Aÿ †_‡K †Qw`Z Ask = 2 f 2 – c = 2 (– 2) 2 – 1 = 2 3 GKK (Ans.) 18. cv‡k¦©i wP‡Î AB Gi Dci PN j¤^| PN Gi •`N ̈© wbY©q Ki| Avevi Ggb GKwU e„‡Ëi mgxKiY wbY©q Ki hv N we›`y‡Z AB †K ̄úk© K‡i| 10 B A 16 P (10, 14) N [BUET 02-03] mgvavb: AB †iLvi mgxKiY, x 10 + y 16 = 1  8x + 5y = 80  PN =    8 × 10 + 5 × 14 – 80| 8 2 + 52 = 70 89 10 B A 16 P (10, 14) N x 10 + y 16 = 1 e„‡Ëi mgxKiY, (x – 10)2 + (y – 14)2 =     70 89 2  89x2 + 89y2 – 1780x – 2492y + 21444 = 0 (Ans.) 19. x 2 + y2 = 45 e„‡Ëi (6, – 3) we›`y‡Z Aw1⁄4Z ̄úk©K x 2 + y2 – 4x + 2y – 35 = 0 e„ˇK A I B we›`y‡Z †Q` K‡i| †`LvI †h, A I B we›`y‡Z ̄úk©KØq ci ̄úi j¤^| [BUET 00-01; RUET 11-12] mgvavb: x 2 + y2 = 45........... (i); e„‡Ëi (6, – 3) we›`y‡Z Aw1⁄4Z ̄úk©‡Ki mgxKiY, 6x – 3y = 45  y = 2x – 15............ (ii) GLb, x 2 + (2x – 15)2 – 30 – 35 = 0  5x2 – 60x + 160 = 0  x = 4, 8  y = – 7, 1  A  (4, – 7) I B  (8, 1) A we›`y‡Z Aw1⁄4Z ̄úk©‡Ki mgxKiY, 4x – 7y – 2(x + 4) + y – 7 – 35 = 0  2x – 6y – 50 = 0 ; m1 = 1 3 B we›`y‡Z Aw1⁄4Z ̄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) 20. (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  wb‡Y©q e„‡Ëi mgxKiY, (x – 2)2 + (y + 1)2 = 1 (Ans.) 21. 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 GKK  e„‡Ëi mgxKiY, (x – 1)2 + (y – 3)2 = 10 (Ans.)