Tiêu đề Circle Practice Sheet HSC FRB 24.pdf ✅

e„Ë  Final Revision Batch '24 1 04 e„Ë The Circle Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 1 1 1 1 2 1 2 1 1 2022 1 2 1 2 1 1 2 1 1 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 4 2 4 3 4 5 5 5 4 2022 4 4 3 3 4 3 5 4 5 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| [XvKv †evW©- Õ23] O C Y X (2, 3) (K) †Kv‡bv e„‡Ëi civwgwZK mgxKiY x 2 = 25 – t 2 , y = t n‡j e„ËwUi e ̈vm wbY©q Ki| (L) Ggb GKwU e„‡Ëi mgxKiY wbY©q Ki hvi †K›`a X A‡ÿi Dci Aew ̄’Z Ges DÏxc‡K DwjøwLZ e„ËwUi †K›`a I g~jwe›`y w`‡q hvq| (M) OC ̄úk©‡Ki mgxKiY wbY©q Ki| DËi: (K) 10 GKK ; (L) 2x2 + 2y2 – 13x = 0 (M) 12x + 5y = 0 2| x 2 + y2 – 2x + 2y = 2 e„‡Ëi GKwU ̄úk©K 3x + 4y – 9 = 0 [ivRkvnx †evW©- Õ23] (K) GKwU e„‡Ëi †K›`a     6   4 Ges e ̈vmva© 5 GKK n‡j, e„‡Ëi mgxKiY wbY©q Ki| (L) DÏxc‡K DwjøwLZ e„‡Ë Giƒc `yBwU ̄úk©‡Ki mgxKiY wbY©q Ki hv DÏxc‡Ki ̄úk©‡Ki Dci j¤^| (M) (4, – 3) we›`y †_‡K DÏxc‡Ki e„ËwUi Dci Aw1⁄4Z ̄úk©‡Ki •`N© ̈ Ges mgxKiY wbY©q Ki| DËi: (K) x 2 + y2 – 6 2x – 6 2y + 11 = 0 (L) 4x – 3y + 3 = 0 ; 4x – 3y – 17 = 0 (M) y + 3 = 0 ; 12x + 5y – 33 = 0 ; 3 GKK 3| `„k ̈Kí-1: x = 0, y = 0 Ges x = 10 wZbwU mij‡iLvi mgxKiY| `„k ̈Kí-2: x 2 + y2 – 12x + 16y – 69 = 0 Ges x 2 + y2 – 9x + 12y – 59 = 0 `yBwU e„‡Ëi mgxKiY| [h‡kvi †evW©- Õ23] (K) (3, 2) we›`y †_‡K 2x2 + 2y2 – 6x – 7 = 0 e„‡Ë Aw1⁄4Z ̄úk©‡Ki •`N© ̈ wbY©q Ki| (L) `„k ̈Kí-1 Gi mij‡iLv wZbwU‡K ̄úk© K‡i Giƒc e„‡Ëi mgxKiY wbY©q Ki| (M) `„k ̈Kí-2 Gi e„Ë `yBwUi mvaviY R ̈v‡K e ̈vm a‡i Aw1⁄4Z e„‡Ëi mgxKiY wbY©q Ki| DËi: (K) 1 2 GKK; (L) x 2 + y2 – 10x  10y + 25 = 0 (M) 5x2 + 5y2 + 12x – 16y – 105 = 0 4| x 2 + y2 + 6x – 6y – 31 = 0 ........ (i) 4x + 3y + 7 = 0 ........ (ii) 2x – 5y + 1 = 0 ........ (iii) [h‡kvi †evW©- Õ23] (K) r = cos – sin e„ËwUi †K›`a wbY©q Ki| (L) (ii) I (iii) bs mij‡iLvi ga ̈eZ©x ̄’~j‡Kv‡Yi mgwØLЇKi mgxKiY wbY©q Ki| (M) (ii) bs †iLvi Dci j¤^ Ges (i) bs e„ˇK ̄úk© K‡i Giƒc mij‡iLvi mgxKiY wbY©q Ki| DËi: (K)     1 2  – 1 2 (L) (4 29 + 10) x + (3 29 – 25) y + 7 29 + 5 = 0 (M) 3x – 4y + 56 = 0 ; 3x – 4y – 14 = 0
2  Higher Math 1st Paper Chapter-4 5| DÏxcK-1: x = 0 6 Y Y X X (0, – 4) DÏxcK-2: x 2 + y2 + 2x + 3y + 1 = 0 GKwU e„‡Ëi mgxKiY| [Kzwgjøv †evW©- Õ23] (K) x 2 + y2 – 12x – 8y + 34 = 0 e„‡Ëi †K›`a I e ̈vmva© wbY©q Ki| (L) DÏxcK-1 Gi e„ËwUi mgxKiY wbY©q Ki| (M) g~jwe›`y (0, 0) †_‡K DÏxcK-2 Gi e„ËwUi Dci Aw1⁄4Z ̄úk©‡Ki mgxKiY I •`N© ̈ wbY©q Ki| DËi: (K) (6, 4) ; 3 2 ; (L) x 2 + y2 – 10x + 8y + 16 = 0 (M) y = 0 ; 12x + 5y = 0 ; 1 GKK 6| e„‡Ëi mgxKiY: x 2 + y2 + 6x + 2y + 6 = 0 x 2 + y2 + 8x + y + 10 = 0 [PÆMÖvg †evW©- Õ23] (K) e ̈vmva© 3 Ges x 2 + y2 – 4x – 6y = 0 e„‡Ëi mv‡_ mg‡Kw›`aK Giƒc e„‡Ëi mgxKiY wbY©q Ki| (L) DÏxc‡K DwjøwLZ e„Ë؇qi mvaviY R ̈v‡K e„‡Ëi e ̈vm a‡i Aw1⁄4Z e„‡Ëi mgxKiY wbY©q Ki| (M) (– 3, 2) we›`y n‡Z DÏxc‡Ki 1g e„ËwUi Dci Aw1⁄4Z ̄úk©K Ges Awfj‡¤^i mgxKiY wbY©q Ki| DËi: (K) x 2 + y2 – 4x – 6y + 4 = 0 (L) 5x2 + 5y2 + 26x + 12y + 22 = 0 (M) 5x – 2y + 3 5 + 4 = 0 ; 5x + 2y + 3 5 – 4 = 0 ; 2x – 5y + 6 + 2 5 = 0 7| [wm‡jU †evW©- Õ23] X X Y  A(– 8, 0) Y O B(0, 6) (K) r – 2cos + 4 sin = 0 e„‡Ëi †K›`a wbY©q Ki| (L) Giƒc e„‡Ëi mgxKiY wbY©q Ki hv y Aÿ‡K B we›`y‡Z ̄úk© K‡i Ges x Aÿ n‡Z AB Gi mgvb •`‡N© ̈i R ̈v KZ©b K‡i| (M) DÏxc‡Ki e„ËwUi mgxKiY wbY©q Ki| DËi: (K) (1, – 2) ; (L) x 2 + y2  2 61x – 12y + 36 = 0 (M) x 2 + y2 + 4x – 4y + 4 = 0 8| [ewikvj †evW©- Õ23] D(0,3) C Y Y X X O A B (K) 2x2 + 2y2 = 0 e„‡Ëi †K‡›`ai ̄’vbv1⁄4 I e ̈vmva© wbY©q Ki| (L) A I B we›`y؇qi ̄’vbv1⁄4 h_vμ‡g (1, 0) I (9, 0) n‡j C †K›`awewkó e„‡Ëi mgxKiY wbY©q Ki| (M) BD Gi mgvšÍivj †iLv DÏxc‡Ki e„ˇK †h we›`y‡Z ̄úk© K‡i Zv wbY©q Ki| DËi: (K) (0, 0) ; 0 GKK ; (L) x 2 + y2 – 10x – 6y + 9 = 0 (M)    10 – 10  2  6 – 3 10 2 9| [ewikvj †evW©- Õ23] (2, 2) Y X X Y O D P 45 E A F B (K) GKwU e„‡Ëi civwgwZK mgxKiY x 2 = 1 – t 2 Ges y = t + 3 n‡j e„ËwUi †K›`a I e ̈vmva© KZ? (L) hw` P we›`ywU EF †iLvs‡ki GKwU mgwÎLÐK nq Z‡e OP †iLvi mgxKiY wbY©q Ki| (M) hw` OD = 3 2 nq Z‡e e„ËwUi mgxKiY wbY©q Ki| DËi: (K) (0, 0) ; 1 GKK (L) 2x – y = 0 ; x – 2y = 0 (M) x 2 + y2 – 4x – 4y + 6 = 0
e„Ë  Final Revision Batch '24 3 10| `„k ̈Kí-1: x 2 + y2 – 6x = 0 ....... (i) x – 4 = 0 ........ (ii) `„k ̈Kí-2: x 2 + y2 + 6x + 4y + 6 = 0 x 2 + y 2 + 4x + 2y + 2 = 0 [w`bvRcyi †evW©- Õ23] (K) (2, – 3) we›`y n‡Z 2x2 + 2y2 = 8 e„‡Ëi Dci Aw1⁄4Z ̄úk©‡Ki •`N© ̈ wbY©q Ki| (L) `„k ̈Kí-1 Gi Av‡jv‡K e„‡Ëi mgxKiY wbY©q Ki hvi †K›`a (7, 0) Ges (i) bs e„Ë Ges (ii) bs †iLvi †Q`we›`y w`‡q hvq| (M) `„k ̈Kí-2 Gi Av‡jv‡K e„Ë `yBwUi mvaviY R ̈v †h e„‡Ëi e ̈vm Zvi mgxKiY wbY©q Ki| DËi: (K) 3 GKK ; (L) x 2 + y2 – 14x + 32 = 0 (M) x 2 + y2 + 3x + y = 0 11| P(1, 2), Q(2, 3) `yBwU we›`y Ges x 2 + y2 – 6x – 4y + 1 = 0 GKwU e„‡Ëi mgxKiY| [gqgbwmsn †evW©- Õ23] (K) 3x2 + 3y2 – 6x – 12y + 1 = 0 e„‡Ëi e ̈vmva© wbY©q Ki| (L) P †K›`awewkó Giƒc e„‡Ëi mgxKiY wbY©q Ki hv cÖ`Ë e„‡Ëi †K›`a w`‡q hvq| (M) P I Q we›`yMvgx Ges y Aÿ‡K ̄úk© K‡i Giƒc e„‡Ëi mgxKiY wbY©q Ki| DËi: (K) 14 3 GKK ; (L) x 2 + y2 – 2x – 4y + 1 = 0 (M) x 2 + y2 – 10x + 2y + 1 = 0 x 2 + y2 – 2x – 6y + 9 = 0 12| `„k ̈Kí-1: f(x, y) = 3x – 4y – 5 Ges g(x, y) = x2 + y2 – 6x + 8y + 9 `„k ̈Kí-2: (5, 3) I (–5, 7) we›`yØq GKwU e„‡Ëi e ̈v‡mi cÖvšÍwe›`y| [XvKv †evW©- Õ22] (K) g(x, y) = 0 e„Ë Øviv y A‡ÿi LwÐZ As‡ki cwigvY wbY©q Ki| (L) †`LvI †h, `„k ̈Kí-1 G ewY©Z f(x, y) = 0 †iLvwU g(x, y) = 0 e„‡Ëi GKwU ̄úk©K| (M) `„k ̈Kí-2 Abyhvqx e„‡Ëi mgxKiY wbY©q Ki| wb‡Y©q e„Ë I f(x, y) = 0 †iLvi †Q`we›`yI g~jwe›`yMvgx e„‡Ëi mgxKiYI wbY©q Ki| DËi: (K) 2 7 GKK ; (M) 5x2 + 5y2 – 12x – 34y = 0 13| [ivRkvnx †evW©- Õ22] A(1, 2) C O Y Y X X D B(3, 2) y = 2x (–10, 0) OD n‡jv e„ËwUi GKwU e ̈vm| (K) r(1 + cos) = 2 mgxKiY‡K Kv‡Z©mxq mgxKi‡Y cÖKvk Ki| (L) OC R ̈v‡K e ̈vm a‡i AswKZ e„‡Ëi mgxKiY wbY©q Ki| (M) A Ges B we›`yMvgx e„Ë x Aÿ‡K ̄úk© Ki‡j Zvi mgxKiY wbY©q Ki| DËi: (K) y 2 + 4x – 4 = 0 ; (L) x 2 + y2 + 2x + 4y = 0 (M) 2x2 + 2y2 – 8x – 5y + 8 = 0 14| [Kzwgjøv †evW©- Õ22] D C(5, 2) 5x + 12y + 16 = 0 (0, 6) O Y X AB = 16 GKK A B (K) (2, 2) we›`y n‡Z x 2 + y2 + 4x – 2y + 4 = 0 e„‡Ë Aw1⁄4Z ̄úk©‡Ki •`N© ̈ wbY©q Ki| (L) C †K›`awewkó e„‡Ëi ̄úk©we›`y D Gi ̄’vbv1⁄4 wbY©q Ki| (M) AB R ̈v wewkó e„‡Ëi mgxKiY wbY©q Ki| DËi: (K) 4 GKK ; (L)     40 13 – 34 13 (M) x 2 + y2 – 20x – 12y + 36 = 0 15| DÏxcK-1: AB mij‡iLvwU cÖ_g PZzf©v‡M 32 3 eM© GKK †ÿÎdj wewkó OAB MVb K‡i Ges g~jwe›`y n‡Z AB Gi Dci j¤^ OP hv x-A‡ÿi abvZ¥K w`‡Ki mv‡_ 60 †Kv‡Y AvbZ| DÏxcK-2: x 2 + y2 + 4x + 4y + 1 = 0 Ges x 2 + y2 + 4x + 3y + 2 = 0 `yBwU e„‡Ëi mgxKiY| [Kzwgjøv †evW©- Õ22] (K) (4, 2) we›`yMvgx Ges 6x + 8y + 17 = 0 †iLvi mv‡_ mgvšÍivj †iLvi mgxKiY wbY©q Ki| (L) DÏxcK-1 G DwjøwLZ AB Gi mgxKiY wbY©q Ki| (M) DÏxcK-2 G DwjøwLZ e„Ë`&¦‡qi mvaviY R ̈v‡K e ̈vm a‡i Aw1⁄4Z e„‡Ëi mgxKiY wbY©q Ki| DËi: (K) 3x + 4y – 20 = 0 ; (L) x + 3y = 8 (M) e„Ë A1⁄4b m¤¢e bq| 16| [h‡kvi †evW©- Õ22] P O Y Y X X C(– 4,2)
4  Higher Math 1st Paper Chapter-4 (K) (1, 3) †K›`awewkó e„Ë Y-Aÿ‡K ̄úk© K‡i| e„ËwUi mgxKiY wbY©q Ki| (L) e„ËwUi GKwU R ̈v Gi mgxKiY wbY©q Ki hvi ga ̈we›`y (– 5, 3) we›`y‡Z Aew ̄’Z| (M) OP ̄úk©‡Ki mgxKiY wbY©q Ki| DËi: (K) x 2 + y2 – 2x – 6y + 9 = 0 ; (L) x – y + 8 = 0 (M) 4x + 3y = 0 17| [PÆMÖvg †evW©- Õ22] X P(2, 1) O B(5, 2) A(3, 0) Y Y X Q(– 6, 5) (K) x 2 + y2 = 121 e„‡Ëi †cvjvi mgxKiY wbY©q Ki| (L) P I Q we›`y؇qi ms‡hvM †iLv‡K e ̈vm a‡i Aw1⁄4Z e„ËwU KZ...©K Aÿ؇qi LwÐZvs‡ki •`N© ̈ wbY©q Ki| (M) ÔOÕ g~jwe›`y †_‡K A I B we›`yMvgx e„‡Ëi Dci Aw1⁄4Z Aci ̄úk©‡Ki mgxKiY wbY©q Ki| DËi: (K) r = 11 (L) x A‡ÿi LwÛZvs‡ki •`N© ̈ 2 11 GKK ; y A‡ÿi LwÛZvs‡ki •`N© ̈ 8 GKK ; (M) 12x – 5y = 0 18| [ewikvj †evW©- Õ22] X 5 2 O Y Y X 3 2 A B (K) OAB Gi †ÿÎdj I Zvi fi‡K›`a wbY©q Ki| (L) AB Gi mgxKiY wbY©q Ki Ges Zvi mgwÎLÐb we›`y؇qi ̄’vbvsK wbY©q Ki| (M) AB †K e ̈vm a‡i AswKZ e„‡Ëi mgxKiY I B we›`y‡Z ̄úk©‡Ki mgxKiY wbY©q Ki| DËi: (K) 15 8 eM© GKK ; =     5 6  1 2 ; (L) x + 10y = 15 ;     5 3  1 2 ;     5 6 1 (M) 2x2 + 2y2 – 5x – 3y = 0 ; 6y – 10x – 9 = 0 19| `„k ̈Kí-1: 3x – 4y + 7 = 0, 4x – 3y + 2 = 0 `„k ̈Kí-2: [ewikvj †evW©- Õ22] X 6 O (0, –2) Y Y X (K) r = b sin 2 †K Kv‡Z©mxq mgxKi‡Y iƒcvšÍi Ki| (L) `„k ̈Kí-1 G ewY©Z mij‡iLv `ywUi ga ̈eZ©x ̄’~j‡Kv‡Yi mgwØLÐK †iLvwU Aÿ؇qi mv‡_ †h wÎfzR Drcbœ K‡i Zvi †ÿÎdj wbY©q Ki| (M) `„k ̈Kí-2 G cÖ`wk©Z e„ËwUi mgxKiY wbY©q Ki| DËi: (K) (x 2 + y 2 ) 3 = 4b2 x 2 y 2 ; (L) 25 2 eM© GKK (M) x 2 + y2 – 2 13x + 4y + 4 = 0 20| [wm‡jU †evW©- Õ22] X A(1,2) O B(3, 2) Y Y X (K) 2x2 + 2y2 + 4x + 6y + 1 = 0 e„ËwU Øviv y A‡ÿi LwÐZ As‡ki •`N© ̈ wbY©q Ki| (L) DÏxc‡Ki Av‡jv‡K e„ËwUi mgxKiY wbY©q Ki| (M) g~jwe›`y †_‡K e„ËwUi Aci ̄úk©‡Ki mgxKiY wbY©q Ki| DËi: (K) 7 ; (L) 2x2 + 2y2 – 8x – 5y + 8 = 0 (M) 80x – 39y = 0 wb‡Y©q ̄úk©‡Ki mgxKiY| 21| x 2 + y2 – 6x + 2y + 1 = 0 Ges x 2 + y2 + 4x + 2y – 4 = 0 `ywU e„‡Ëi mgxKiY| [w`bvRcyi †evW©- Õ22] (K) r = 4 cos e„ËwUi †K‡›`ai Kv‡Z©mxq ̄’vbvsK wbY©q Ki| (L) DÏxc‡K ewY©Z cÖ_g e„‡Ëi GKwU ̄úk©‡Ki mgxKiY wbY©q Ki hv 3x + 4y – 1 = 0 Gi mgvšÍivj| (M) DÏxc‡K ewY©Z e„Ë؇qi mvaviY R ̈v †h e„‡Ëi e ̈vm Zvi mgxKiY wbY©q Ki| DËi: (K) (2 ,0) ; (L) 3x + 4y + 10 = 0 ; 3x + 4y – 20 = 0 (M) 2(x2 + y2 ) – 2x + 4y – 3 = 0