Tiêu đề Circle Practice Sheet Solution Varsity.pdf ✅

e„Ë  Varsity Practice Sheet 1 e„Ë The Circle PZz_© Aa ̈vq 1. ax2 + by2 + c = 0 mgxKiYwU †Kvb k‡Z© GKwU e„Ë wb‡`©k Ki‡e? c = r2 c = 0 a = b, b  0 a  b, a = 0 DËi: a = b, b  0 e ̈vL ̈v: e„‡Ëi mgxKi‡Yi x 2 Ges y 2 Gi mnM mgvb A_©vr a = b, b  0 2. wb‡Pi †KvbwU e„‡Ëi mgxKiY? 3x2 + 4y2 + 4x + 6y + 8 = 0 x 2 + y2 + 3x + 6y + 8 = 0 3x2 + 3y2 + 5xy + 4x + 2 = 0 x 2 + y3 + 2x + 4y + 2 = 0 DËi: x 2 + y2 + 3x + 6y + 8 = 0 e ̈vL ̈v: e„‡Ëi mgxKi‡Yi x 2 Ges y 2 Gi mnM mgvb A_©vr a = b, a  0, b  0 Ges xy hy3 c` _vK‡e bv| 3. k (x + 1) 2 4 + (y – 1) 2 3 = 0 GKwU e„‡Ëi mgxKiY wb‡`©k Ki‡j k Gi gvb KZ? 3 2 1 3 4 4 3 DËi: 4 3 e ̈vL ̈v: k 4 = 1 3  k = 4 3 [e„Ë nevi kZ©: x 2 I y 2 Gi mnM mgvb n‡e Ges xy hy3 c` _vK‡e bv|] 4. 2x2 + 2y2 + 4x – 2y + 4 = 0 e„‡Ëi †K›`a †KvbwU?     – 1  1 2     1  – 1 2 (– 2, 1)     2  – 1 2 DËi:     – 1  1 2 e ̈vL ̈v: 2x2 + 2y2 + 4x – 2y + 4 = 0  x 2 + y2 + 2x – y + 2 = 0  †K›`a     – 1  1 2 5. 3x2 + 3y2 + 6x – 12y – 15 = 0 mgxKiY Øviv ewY©Z e„‡Ëi †K›`a Kx? (– 3, 6) (– 1, 2) (– 3, 2) (– 3, 12) DËi: (– 1, 2) e ̈vL ̈v: 3x2 + 3y2 + 6x – 12y – 15 = 0  x 2 + y2 + 2x – 4y – 5 = 0  †K›`a (– 1, 2) 6. x 2 + y2 – 6x – 8y – 75 = 0 e„‡Ëi e ̈mva© KZ? 100 – 50 10 20 DËi: 10 e ̈vL ̈v: r = g 2 + f2 – c = (–3) 2 + (– 4) 2 – (– 75) = 10 7. 2x2 + 2y2 – 4x – 12y + 11 = 0 e„ËwUi e ̈vmva© KZ? 29 31 2 29 9 2 DËi: 9 2 e ̈vL ̈v: 2x2 + 2y2 – 4x – 12y + 11 = 0  x 2 + y2 – 2x – 6y + 11 2 = 0  e ̈vmva© = 1 + 9 – 11 2 = 9 2 8. r 2 + 2rsin = 3 e„ËwUi †K›`a KZ? (1, 0) (– 1, 0) (0, 1) (0, – 1) DËi: (0, – 1) e ̈vL ̈v: r 2 + 2rsin = 3  x 2 + y2 + 2y – 3 = 0  †K›`a (0, – 1) 9. r = 2cos †cvjvi mgxKiYwU wb‡`©k K‡iÑ mij‡iLv cive„Ë Dce„Ë e„Ë DËi: e„Ë e ̈vL ̈v: r = 2cos  r 2 = 2rcos  x 2 + y2 = 2x  x 2 + y2 – 2x = 0 hv GKwU e„‡Ëi mgxKiY|
2  Higher Math 1st Paper Chapter-4 10. (– 4, 3) Ges (12, – 1) we›`y؇qi ms‡hvM †iLvsk‡K e ̈vm a‡i Aw1⁄4Z e„‡Ëi mgxKiY †KvbwU? x 2 + y2 + 8x – 2y + 51 = 0 x 2 + y2 – 8x – 2y + 51 = 0 x 2 + y2 + 8x + 2y – 51 = 0 x 2 + y2 – 8x – 2y – 51 = 0 DËi: x 2 + y2 – 8x – 2y – 51 = 0 e ̈vL ̈v: (– 4, 3) Ges (12, – 1) †K e ̈vm a‡i e„‡Ëi mgxKiY, (x + 4) (x – 12) + (y – 3) (y + 1) = 0  x 2 – 8x – 48 + y2 – 2y – 3 = 0  x 2 + y2 – 8x – 2y – 51 = 0 11. (3, – 1) we›`yMvgx Ges x 2 + y2 – 6x + 8y = 0 e„‡Ëi mv‡_ GK‡Kw›`aK e„‡Ëi mgxKiYÑ x 2 + y2 + 6x – 8y + 16 = 0 x 2 + y2 – 6x – 8y – 16 = 0 x 2 + y2 – 6x + 8y + 16 = 0 x 2 + y2 – 6x – 8y + 16 = 0 DËi: x 2 + y2 – 6x + 8y + 16 = 0 e ̈vL ̈v: x 2 + y2 – 6x + 8y = 0 e„‡Ëi †K›`a (3, – 4) Ges wb‡Y©q e„‡Ëi e ̈vmva© (3 – 3) 2 + (– 4 + 1) 2 = 3  e„‡Ëi mgxKiY (x – 3)2 + (y + 4)2 = 32  x 2 + y2 – 6x + 8y + 16 = 0 12. GKwU e„‡Ëi mgxKiY wbY©q Ki hvi †K‡›`ai ̄’vbv1⁄4 (2, 3) Ges x + y – 2 = 0 †iLvwU e„ˇK ̄úk© K‡i| 2 (x2 + y2 ) – 8x – 12y + 17 = 0 2 (x2 + y2 ) – 6x – 10y + 15 = 0 2 (x2 + y2 ) – 4x – 8y + 11 = 0 2 (x2 + y2 ) – 2x – 6y + 7 = 0 DËi: 2 (x2 + y2 ) – 8x – 12y + 17 = 0 e ̈vL ̈v: †h‡nZz wb‡Y©q e„‡Ëi †K‡›`ai ̄’vbv1⁄4 (2, 3) Ges ̄úk©K x + y – 2 = 0 †m‡nZz wb‡Y©q e„‡Ëi e ̈vmva© =     2 + 3 – 2 1 2 + 12 = 3 2  e„‡Ëi mgxKiY, (x – 2)2 + (y – 3)2 =     3 2 2  x 2 – 4x + 4 + y2 – 6y + 9 = 9 2  2 (x2 + y2 ) – 8x – 12y + 17 = 0 13. k Gi †Kvb gv‡bi Rb ̈ (x – y + 3)2 + (kx + 2) (y – 1) = 0 mgxKiYwU GKwU e„Ë wb‡`©k K‡i? 2 1 – 2 – 1 DËi: 2 e ̈vL ̈v: (x – y + 3)2 + (kx + 2) (y – 1) = 0 x 2 + y2 + 9 – 2xy + 6x – 6y + kxy – kx + 2y – 2 = 0  x 2 + y2 + (k – 2) xy + (6 – k) x – 4y + 7 = 0 mgxKiYwU e„Ë n‡Z n‡j xy hy3 c` _vK‡Z cvi‡e bv A_©vr k – 2 = 0  k = 2 14. x 2 + y2 – 5x = 0, x2 + y2 + 3x = 0 e„Ë؇qi †K‡›`ai `~iZ¡ KZ? 4 units 1 units 5 units 2 units DËi: 4 units e ̈vL ̈v: x 2 + y2 – 5x = 0 e„‡Ëi †K›`a     5 2  0 x 2 + y2 + 3x = 0 e„‡Ëi †K›`a    –  3 2  0 †K›`a؇qi ga ̈eZ©x `~iZ¡ =     5 2 + 3 2 2 = 4 15. x 2 + y2 – 6x = 0 Ges x 2 + y2 – 8y = 0 e„Ë؇qi †K›`a `ywUi ga ̈eZ©x `~iZ¡ KZ? 1 5 7 10 DËi: 5 e ̈vL ̈v: x 2 + y2 – 6x = 0 Gi †K›`a (3, 0) Ges x 2 + y2 – 8y = 0 Gi †K›`a (0, 4) †K›`a؇qi ga ̈eZ©x `~iZ¡ = (3 – 0) 2 + (0 – 4) 2 = 5 16. (5, 0) Ges (0, 5) we›`y‡Z Aÿ‡iLvØq‡K ̄úk©Kvix e„‡Ëi mgxKiYÑ x 2 + y2 + 10x – 10y – 25 = 0 x 2 + y2 + 10x + 10y + 25 = 0 x 2 + y 2 – 10x + 10y – 25 = 0 x 2 + y2 – 10x – 10y + 25 = 0 DËi: x 2 + y2 – 10x – 10y + 25 = 0 e ̈vL ̈v: g‡b Kwi, e„‡Ëi mgxKiY x 2 + y2 + 2gx + 2fy + c = 0 †h‡nZz e„ËwU (5, 0) Ges (0, 5) we›`y‡Z Aÿ †iLvØq‡K ̄úk© K‡i|  e„‡Ëi mgxKiY x 2 + y2 – 10x – 10y + 25 = 0 17. x 2 + y2 – gx = 0 e„Ë Øviv Ave× †ÿ‡Îi †ÿÎdj KZ eM© GKK? 1 8 g 2 1 4 g 2 1 2 g 2 g 2 DËi: 1 4 g 2 e ̈vL ̈v: x 2 + y2 + 2.    –  g 2 .x + 2.0.y + 0 = 0 e ̈vmva© =    –  g 2 2 = g 2  †ÿÎdj = r 2 = 1 4 g 2 18. GKK e ̈vmv‡a©i e„‡Ë AšÍwj©wLZ GKwU mgevû wÎfz‡Ri evûi •`N© ̈Ñ 1 2 units 3 2 units 3 units 1 units
e„Ë  Varsity Practice Sheet 3 DËi: 3 units e ̈vL ̈v: sin60 = a 2R  a = 2Rsin60  a = 2.1. 3 2 = 3 19. 2x2 + 2y2 + 6x + 10y – c = 0 e„‡Ëi e ̈vmva© 3 n‡j c Gi gvb KZ? 1 – 1 2 – 2 DËi: 1 e ̈vL ̈v: 2x2 + 2y2 + 6x + 10y – c = 0  x 2 + y2 + 3x + 5y – c 2 = 0 g = 3 2 , f = 5 2 Ges, c = – c 2 e ̈vmva© = g 2 + f2 – c = 3      3 2 2 +     5 2 2 + c 2 = 32  c 2 = 9 – 9 4 – 25 4 = 36 – 9 – 25 4 = 2 4  c 2 = 1 2  c = 1 20. x 2 + y2 – by = 0 e„‡Ëi mgxKiY †cvjvi ̄’vbv‡1⁄4i gva ̈‡g cÖKvk Ki‡j mgxKiYwU n‡eÑ x = ysin r = b r = bsin r = bcos DËi: r = bsin e ̈vL ̈v: (x2 + y2 ) – by = 0  r 2 – brsin = 0  r = bsin 21. x Aÿ‡K (2, 0) we›`y‡Z ̄úk© K‡i Ges (– 1, 9) we›`y w`‡q hvq Giƒc e„‡Ëi mgxKiY x 2 + y2 – 4x – 10y + 4 = 0 x 2 + 4y2 – x – 10y + 4 = 0 4x2 + y2 – 4x – 10y = 0 x 2 + y2 – 4x – 6y + 86 = 0 DËi: x 2 + y2 – 4x – 10y + 4 = 0 e ̈vL ̈v: I e„Ë bq| m¤¢e DËi A_ev | (2, 0) we›`yMvgx bq| wKš‘ (2, 0) we›`yMvgx ZvB DËi n‡e| 22. x 2 + y2 + 4x + 2fy + c = 0 e„Ë g~jwe›`y‡Z y Aÿ‡K ̄úk© Ki‡j f Gi gvb wbY©q Ki? – 2 2 1 0 DËi: 0 e ̈vL ̈v: g~jwe›`yMvgx n‡j, c = 0| e„ËwU y Aÿ‡K ̄úk© Ki‡j f 2 = c  f = 0 23. (3, 4) †K›`awewkó GKwU e„Ë x Aÿ‡K ̄úk© Ki‡j Dnvi e ̈vmva© KZ? 3 GKK 4 GKK 5 GKK 7 GKK DËi: 4 GKK e ̈vL ̈v: e ̈vmva© = |†K‡›`ai †KvwU | = 4 GKK Note: x Aÿ‡K ̄úk© Ki‡j e ̈vmva© = |†K‡›`ai †KvwU | y Aÿ‡K ̄úk© Ki‡j e ̈vmva© = |†K‡›`ai fzR| 24. (– 2, 1) †K›`awewkó e„Ë x Aÿ‡K ̄úk© Ki‡j e„‡Ëi e ̈vmÑ – 2 1 2 4 DËi: 2 e ̈vL ̈v: e ̈vmva© = | †K‡›`ai †KvwU | = 1 GKK  e ̈vm = 2  e ̈vmva© = 2  1 = 2 25. (5, 4) †K›`awewkó GKwU e„Ë y Aÿ‡K ̄úk© Ki‡j e„ËwUi e ̈vmva© KZ? 5 4 41 9 DËi: 5 e ̈vL ̈v: e ̈vmva© = | †K‡›`ai fzR | = 5 GKK Note: x Aÿ‡K ̄úk© Ki‡j e ̈vmva© = |†K‡›`ai †KvwU | y Aÿ‡K ̄úk© Ki‡j e ̈vmva© = |†K‡›`ai fzR| 26. (– 7, 8) †K›`awewkó e„Ë y Aÿ‡K ̄úk© Ki‡j e„ËwUi e ̈vm KZ? 7 8 14 16 DËi: 14 e ̈vL ̈v: e ̈vmva© = | †K‡›`ai fzR | = 7 GKK  e ̈vm = 2  e ̈vmva© = 2  7 = 14 27. x 2 + y2 + 4x – 6y – 12 = 0 e„Ë Øviv x A‡ÿi LwÐZvs‡ki cwigvY KZ? 4 21 8 2 21 DËi: 8 e ̈vL ̈v: x A‡ÿi LwÐZvsk = 2 g 2 – c = 2 2 2 – (– 12) = 8 28. x 2 + y2 + 2x + 4y – 1 = 0 e„Ë Øviv y A‡ÿi LwÐZvs‡ki •`N© ̈ KZ GKK? 2 2 5 2 5 2
4  Higher Math 1st Paper Chapter-4 DËi: 2 5 e ̈vL ̈v: x A‡ÿi LwÐZvsk = 2 f 2 – c = 2 2 2 – (– 1) = 2 5 29. (2, 3) †K›`awewkó GKwU e„Ë x Aÿ‡K ̄úk© K‡i| e„ËwU Øviv y A‡ÿi LwÐZvs‡ki cwigvY KZ GKK| 5 2 5 2 5 5 DËi: 2 5 e ̈vL ̈v: (x – 2)2 + (y – 3)2 = 32  x 2 + y2 – 4x – 6y + 4 = 0 y A‡ÿi LwÐZvsk = 2 f 2 – c = 2 (– 3) 2 – 4 = 2 5 30. we›`y e„‡Ëi mgxKiYÑ x 2 – y 2 = 0 x 2 + y2 = 0 x 2 + y2 = r2 x 2 + y2 + x + y + 1 = 0 DËi: x 2 + y2 = 0 e ̈vL ̈v: †h e„‡Ëi e ̈vmva© = 0 GKK ZvB we›`ye„Ë| Option G e„‡Ëi e ̈vmva© = 0 31. x 2 + y2 = 0 Kx‡mi mgxKiY? mij‡iLvi mgxKiY we›`y e„‡Ëi mgxKiY Dce„‡Ëi mgxKiY †Kv‡bvwUB bq DËi: we›`y e„‡Ëi mgxKiY e ̈vL ̈v: x 2 + y2 = 0  x 2 + y2 = 02 GLv‡b e ̈vmva© = 0  GwU GKwU we›`y e„Ë| 32. wb‡Pi †KvbwU we›`y e„‡Ëi mgxKiY? x 2 + y2 + 4x + 8y – 20 = 0 x 2 + y2 + 4x – 8y + 20 = 0 x 2 + 4y2 + y2 = 0 x 2 – 2x + y2 = 0 DËi: x 2 + y2 + 4x – 8y + 20 = 0 e ̈vL ̈v: †h e„‡Ëi e ̈vmva© k~b ̈ Zv‡K we›`y e„Ë e‡j| Ackb G x 2 + y2 + 4x – 8y + 20 = 0 e„‡Ëi e ̈vmva© = g 2 + f2 – c = 2 2 + (– 4) 2 – 20 = 20 – 20 = 0 33. 4 (x2 + y2 ) = 5 e„‡Ëi †ÿÎdj KZ eM© GKK?  4 5 4 15 4 25 4 DËi: 5 4 e ̈vL ̈v: †ÿÎdj = r 2 =   5 4 = 5 4 34. x 2 – 6x + y2 + 8y = 0 eμ‡iLv Øviv mxgve× †ÿ‡Î †ÿÎdj n‡eÑ 25 r 2 2 2 75 DËi: 25 e ̈vL ̈v: †ÿÎdj = r 2 =  (5)2 = 25 r = g 2 + f2 – c = (– 3) 2 + (4) 2 – 0 = 9 + 16 = 5 35. (2, –3) †K›`awewkó e„ËwU x-Aÿ‡K ̄úk© Ki‡j Zvi mgxKiY †KvbwU? (x – 2)2 + (y – 3)2 = 32 (x + 2)2 + (y + 3)2 = 22 (x + 2)2 + (y – 3)2 = 22 (x – 2)2 + (y + 3)2 = 32 DËi: (x – 2)2 + (y + 3)2 = 32 e ̈vL ̈v: x Aÿ‡K ̄úk© Ki‡j e ̈vmva© = | †K‡›`ai †KvwU | = 3 e„‡Ëi mgxKiY, (x – 2)2 + (y + 3)2 = 32 36. (3, 5) †K›`awewkó I y Aÿ‡K ̄úk© K‡i GiKg e„‡Ëi mgxKiYÑ x 2 + y2 – 6x – 10y + 25 = 0 x 2 + y2 – 10x – 6y + 25 = 0 x 2 + y2 – 6x – 10y + 9 = 0 x 2 + y2 – 10x – 6y + 9 = 0 DËi: x 2 + y2 – 6x – 10y + 25 = 0 e ̈vL ̈v: (x – 3)2 + (y – 5)2 = 3 2  x 2 + y2 – 6x – 10y + 9 + 25 = 9  x 2 + y2 – 6x – 10y + 25 = 0 37. x 2 + y2 + 2gx + 2fy + c = 0 e„ËwU x Aÿ‡K †Q` K‡i bv, hLbÑ g 2 > c g 2 < c f 2 > c f 2 < c DËi: g 2 < c e ̈vL ̈v: g 2 c n‡j x Aÿ‡K ̄úk© K‡i| g 2 c n‡j x Aÿ‡K ̄úk©/†Q` K‡i bv| g 2 c n‡j x Aÿ‡K †Q` K‡i| f 2 c n‡j y Aÿ‡K ̄úk© K‡i| f 2 c n‡j y Aÿ‡K ̄úk©/†Q` K‡i bv| f 2 c n‡j y Aÿ‡K †Q` K‡i|