Title Circle Varsity Practice Sheet Solution.pdf ✅

e„Ë  Varsity Practice Sheet Solution 1 04 e„Ë The Circle weMZ mv‡j DU-G Avmv cÖkœvejx 1. r = sin e„‡Ëi †K›`a I e ̈vmva© KZ n‡e? [DU 23-24]     1 2  0 , 1 2 (0, 2) , 1 2     0  1 2 , 2     0  1 2 , 1 2 DËi:     0  1 2 , 1 2 e ̈vL ̈v: r = sin  r 2 – rsin = 0  x 2 + y2 – y = 0  g = 0 ; f = – 1 2 ; c = 0  †K›`a     0  1 2  e ̈vmva© = 1 2 GKK 2. 3x + 4y = k †iLvwU x 2 + y2 = 10x e„ˇK ̄úk© K‡i| k Gi gvb KZ? [DU 21-22; CU 20-21] 8, – 30 – 8, 30 – 10, 40 10, – 40 DËi: – 10, 40 e ̈vL ̈v: x 2 + y2 – 10x = 0 e„‡Ëi †K›`a (5, 0) Ges e ̈vmva© = 5 2 + 0 – 0 = 5 GKK ̄úk©‡Ki mgxKiY: 3x + 4y – k = 0 †K›`a n‡Z ̄úk©‡Ki j¤^ `~iZ¡ e ̈vmv‡a©i mgvb n‡e|      3  5 + 4  0 – k 3 2 + 42 = 5  |15 – k| = 5  5  15 – k =  25  k = – 10, 40 3. x = acos + bsin, y = asin – bcos †Kvb KwY‡Ki mgxKiY? [DU 20-21; CU 22-23] ellipse parabola circle hyperbola DËi: circle e ̈vL ̈v: x = acos + bsin ......... (i) y = asin – bcos ......... (ii) (i)2 + (ii)2 K‡i, x 2 + y2 = a2 cos2  + b2 sin2  + 2absin.cos + a2 sin2  + b 2 cos2  – 2absin.cos  x 2 + y2 = a2 (sin2  + cos2 ) + b2 (sin2  + cos2 )  x 2 + y2 = a2 + b2 ; hv GKwU e„‡Ëi mgxKiY wb‡`©k K‡i| 4. †cvjvi ̄’vbv1⁄4 r 2 – 2rsin = 3 GKwU e„‡Ëi mgxKiY| e„ËwUi e ̈vmva© n‡eÑ [DU 20-21] 2 3 4 6 DËi: 2 e ̈vL ̈v: GLv‡b, r 2 – 2rsin = 3  x 2 + y2 – 2y – 3 = 0  g = 0; f = – 2 2 = – 1; c = – 3  e ̈vmva©, r = g 2 + f 2 – c = 0 2 + 1 + 3 = 4 = 2 GKK 5. (4, 3) †K›`a wewkó Ges 5x – 12y + 3 = 0 mij‡iLv‡K ̄úk© K‡i Ggb e„‡Ëi mgxKiY †KvbwU? [DU 19-20] x 2 + y2 + 8x – 6y + 24 = 0 x 2 + y2 – 8x – 6y + 24 = 0 x 2 + y2 + 8x + 6y + 24 = 0 x 2 + y2 – 8x – 6y – 24 = 0 DËi: x 2 + y2 – 8x – 6y + 24 = 0 e ̈vL ̈v: e„‡Ëi †K›`a (4, 3) Ges ̄úk©‡Ki mgxKiY, 5x – 12y + 3 = 0 GLb, †K›`a †_‡K ̄úk©‡Ki j¤^ `~iZ¡ n‡e e ̈vmv‡a©i mgvb|      20 – 36 + 3 5 2 + 122 = r  r =     – 13 169 =     – 13 13 = 1  e„‡Ëi mgxKiY (x – 4)2 + (y – 3)2 = 12  x 2 + y2 – 8x – 6y + 16 + 9 = 1  x 2 + y2 – 8x – 6y + 24 = 0
2  Higher Math 1st Paper Chapter-4 6. x 2 + y2 + 2x – 4y + 4 = 0 e„‡Ëi GKwU ̄úk©KÑ [DU 18-19] x = 0 x = 2 y = 2 y = 4 DËi: x = 0 e ̈vL ̈v: x 2 + y2 + 2x – 4y + 4 = 0 G e„‡Ë, g = 1; f = – 4 2 = – 2; c = 4 f 2 = (– 2)2 = 4 = c  e„ËwU y Aÿ‡K ̄úk© Ki‡e| y A‡ÿi mgxKiY, x = 0 7. 3x2 + 3y2 – 5x – 6y + 4 = 0 e„ËwUi †K›`aÑ [DU 18-19]     1  3 2     5 6  1     5 3  1     2 3  – 1 DËi:     5 6  1 e ̈vL ̈v: 3x2 + 3y2 – 5x – 6y + 4 = 0  x 2 + y2 – 5 3 x – 2y + 4 3 = 0 ........ (i) (i) bs e„‡Ëi †K›`a (– g, – f)       –  5 3 – 2  – 2 – 2      5 6  1 8. g~jwe›`yMvgx GKwU e„Ë abvZ¥K x Aÿ n‡Z 4 GKK Ges abvZ¥K y Aÿ n‡Z 2 GKK Ask KZ©b Ki‡j, Gi mgxKiY n‡eÑ [DU 17-18] x 2 + y2 – 4x – 2y = 0 x 2 + y2 + 4x + 2y = 0 x 2 + y2 + 2x + 4y = 0 x 2 + y2 – 2x – 4y = 0 DËi: x 2 + y2 – 4x – 2y = 0 e ̈vL ̈v: g~jwe›`yMvgx †Kv‡bv e„Ë x Aÿ n‡Z a I y Aÿ n‡Z b GKK Ask KZ©b Ki‡j, Gi mgxKiY n‡e, x 2 + y2 – ax – by = 0  wb‡Y©q mgxKiY: x 2 + y2 – 4x – 2y = 0 Shortcut: g~jwe›`yMvgx †Kv‡bv e„Ë x Aÿ n‡Z a I y Aÿ n‡Z b GKK Ask KZ©b Ki‡j, Gi mgxKiY n‡e, x 2 + y2 – ax – by = 0 9. y Aÿ‡K (0, 4) we›`y‡Z ̄úk© K‡i Ges †K›`a 5x – 7y – 2 = 0 †iLvi Dci Aew ̄’Z e„‡Ëi mgxKiY n‡eÑ [DU 16-17] x 2 + y2 + 12 – 8y + 16 = 0 x 2 + y2 – 8x – 6y + 8 = 0 x 2 + y2 – 12x – 8y + 16 = 0 x 2 + y2 + 8x + 6y – 40 = 0 DËi: x 2 + y2 – 12x – 8y + 16 = 0 e ̈vL ̈v: Option Test: cÖ_‡g e„Ë ̧‡jvi †K›`a †ei Kwi| ïay bs e„‡Ëi †K›`a Øviv DÏxc‡K D‡jøwLZ mij‡iLvi mgxKiY wm× nq|  mwVK DËi 10. (x – 2)2 + (y – 3)2 = 16 Ges (x – 2)2 + (y – 10)2 = 9 e„Ë؇qi ̄úk©we›`yi ̄’vbv1⁄4Ñ [DU 15-16] (2, 3) (16, 9) (2, 10) (2, 7) DËi: (2, 7) e ̈vL ̈v: Option Test: ̄úk©we›`ywU Dfq e„ˇKB wm× Ki‡e| A_ev, (x – 2)2 + (y – 3)2 = 16 e„‡Ëi †K›`a C1(2, 3); e ̈vmva©, r1 = 4 (x – 2)2 + (y –) 2 = 9 e„‡Ëi †K›`a C2(2, 10); e ̈vmva©, r2 = 3 awi, ̄úk©we›`yi ̄’vbv1⁄4 (x, y) (x, y) we›`y C1C2 †K 4 : 3 Abycv‡Z AšÍwe©f3 K‡i|  x = 4  2 + 3  2 4 + 3 = 2  y = 4  10 + 3  3 4 + 3 = 7  ̄úk©we›`yi ̄’vbv1⁄4 (2, 7) 11. ( – 4, 3) Ges (12, – 1) we›`y؇qi ms‡hvM †iLvsk‡K e ̈vm a‡i Aw1⁄4Z e„‡Ëi mgxKiYÑ [DU 15-16] 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) I (12, – 1) we›`y؇qi ms‡hvM †iLvsk‡K e ̈vm a‡i Aw1⁄4Z e„‡Ëi mgxKiY, (x + 4)(x – 12) + (y – 3)(y + 1) = 0  x 2 – 12x + 4x – 48 + y2 + y – 3y – 3 = 0  x 2 + y2 – 8x – 2y – 51 = 0 12. wb‡¤œi †Kvb e„ËwU x Aÿ‡K ̄úk© K‡i? [DU 14-15] x 2 + y2 – 2x + 6y + 4 = 0 x 2 + y2 – 4x + 6y + 5 = 0 x 2 + y2 – 2x + 6y + 1 = 0 2x 2 + 2y 2 – 2x + 6y + 3 = 0 DËi: x 2 + y2 – 2x + 6y + 1 = 0 e ̈vL ̈v: †h AckbwU g 2 = c kZ© †g‡b Pj‡e †mwUB n‡e DËi| bs e„‡Ëi mgxKi‡Y, g = – 2 2 = – 1 Ges c = 1  c = g2 = 1  bs AckbwU g 2 = c kZ© †g‡b P‡j| ZvB GwUB n‡e DËi| 13. (3, – 1) we›`yMvgx Ges x 2 + y2 – 6x + 8y = 0 e„‡Ëi mv‡_ GK‡Kw›`aK e„‡Ëi mgxKiY †KvbwU? [DU 13-14] 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: D3 e„‡Ëi mv‡_ GK‡Kw›`aK e„‡Ëi mgxKiY, x 2 + y2 – 6x + 8y + c = 0, hv (3, – 1) we›`yMvgx|  3 2 + (– 1)2 – 6  3 + 8(– 1) + c = 0  c = 16  e„ËwU x 2 + y2 – 6x + 8y + 16 = 0
e„Ë  Varsity Practice Sheet Solution 3 weMZ mv‡j GST-G Avmv cÖkœvejx 1. x 2 + y2 = 9 e„‡Ëi ̄úk©K x A‡ÿi mv‡_ 45 †KvY Drcbœ K‡i, ̄úk©‡Ki mgxKiY †KvbwU? [GST 23-24; RU 22-23] x + y  3 2 = 0 x – y  3 2 = 0 x + y  2 3 = 0 x – y  2 3 = 0 DËi: x – y  3 2 = 0 e ̈vL ̈v: ̄úk©‡Ki Xvj = tan45 = 1 awi, mgxKiY y = mx + c  x – y + c = 0 e„‡Ëi †K›`a C(0, 0); e ̈vmva©, r = 3 †K›`a †_‡K ̄úk©‡Ki j¤^ `~iZ¡ e ̈vmv‡a©i mgvb|     c 1 + 1 = 3  c =  3 2  mgxKiY, x – y  3 2 = 0 A_ev, y = mx + c, x2 + y2 = r2 Gi ̄úk©‡Ki mgxKiY n‡j, c =  r 1 + m2 m = tan45 = 1  y = mx  r 1 + m2  y = x  3 1 + 12  x – y  3 2 = 0 2. (x – 3)2 + (y – 2)2 = 25 e„‡Ëi GKwU R ̈v †K‡›`a  2 †KvY •Zwi K‡i| R ̈v-wUi •`N© ̈ KZ GKK? [GST 22-23] 5 3 5 3 2 5 2 7 3 DËi: 5 2 e ̈vL ̈v: (x – 3)2 + (y – 2)2 = 25 e„‡Ëi e ̈vmva© 5 GKK 5 5 x  2 wc_v‡Mviv‡mi Dccv` ̈ Abymv‡i cvB, x 2 = 52 + 52  x 2 = 50  x = 50 = 5 2 GKK 3. r = 8cos + 6sin KwYK Øviv x A‡ÿi LwÛZ As‡ki •`N© ̈ KZ GKK? [GST 22-23] 8 6 4 3 DËi: 8 e ̈vL ̈v: r = 8cos + 6sin  r 2 = 8rcos + 6rsin  x 2 + y2 = 8x + 6y  x 2 + y2 – 8x – 6y = 0 GLv‡b, g = – 8 2 = – 4; c = 0  x A‡ÿi LwÛZvsk = 2 g 2 – c = 2 (– 4) 2 – 0 = 2  4 = 8 GKK 4. (4, 3) †K›`awewkó e„‡Ëi e ̈v‡mi GK cÖv‡šÍi ̄’vbv1⁄4 (3, 1) n‡j, Aci cÖv‡šÍi ̄’vbv1⁄4 †KvbwU? [GST 21-22] (4, 0) (5, – 5) (4, 7) (5, 5) DËi: (5, 5) e ̈vL ̈v: awi, e ̈v‡mi Aci cÖv‡šÍi ̄’vbv1⁄4 (, ) GLv‡b, †K›`a (4, 3) n‡jv e ̈v‡mi `yB cÖvšÍ (3, 1) I (, ) Gi ga ̈we›`y|  4 = 3 +  2   = 5  3 = 1 +  2   = 5  Aci cÖv‡šÍi ̄’vbv1⁄4 (5, 5) 5. 3 GKK •`‡N© ̈i GKwU R ̈v e„‡Ëi †K‡›`a  3 †KvY Drcbœ Ki‡j e„‡Ëi †ÿÎdj KZ eM© GKK? [GST 20-21]  3  3 3 DËi: 3 e ̈vL ̈v: R ̈v, AB = 3  AC = AB 2 = 3 2 e ̈vmva©, OA = a A B  6 O 3/2 C a AOC G, sin  6 = AC OA = 3 2 a  a = 3 GKK  †ÿÎdj = a 2 = 3 eM© GKK
4  Higher Math 1st Paper Chapter-4 weMZ mv‡j Agri-G Avmv cÖkœvejx 1. GKwU e„‡Ëi †K›`a (4, 3) hv x 2 + y2 = 9 e„ˇK ewnt ̄’fv‡e ̄úk© K‡i, e„ËwUi e ̈vmva© KZ? [Agri. Guccho 21-22] 2 5 8 3 DËi: 2 e ̈vL ̈v: x 2 + y2 = 9 e„‡Ëi †K›`a C1(0,0) Ges e ̈vmva©, r1 = 3 bZzb e„‡Ëi †K›`a C2(4, 3) Ges e ̈vmva©, r2 = ? Avgiv Rvwb, e„ËØq ci ̄úi ewnt ̄’fv‡e ̄úk© Ki‡j, C1C2 = r1 + r2  (3 – 0) 2 + (4 – 0) 2 = 3 + r2  r2 = 5 – 3  r2 = 2 2. k Gi †Kvb gv‡bi Rb ̈ (x – y + 3)2 + (kx + 2)(y – 1) = 0 mgxKiYwU GKwU e„Ë n‡e? [Agri. Guccho 20-21; RU 23-24; JKKNIU 19-20; SUST 11-12] 1 – 1 – 2 2 DËi: 2 e ̈vL ̈v: (x – y + 3)2 + (kx + 2)(y – 1) = 0  x 2 + y2 + 9 – 2xy – 6y + 6x + kxy – kx + 2y – 2 = 0 e„‡Ëi mgxKi‡Y xy m¤^wjZ †Kv‡bv c` _v‡K bv|  – 2xy + kxy = 0  – 2 + k = 0  k = 2 3. hw` x 2 + y2 + 2gx + 2fy + c = 0 e„ËwU y Aÿ‡K ̄úk© K‡i Z‡eÑ [Agri. Guccho 19-20] g 2 = c f 2 = c g 2 – f 2 = c g 2 + f 2 = c DËi: f 2 = c e ̈vL ̈v: †Kv‡bv e„Ë x Aÿ‡K ̄úk© Ki‡j g 2 = c Ges y Aÿ‡K ̄úk© Ki‡j f 2 = c nq| weMZ mv‡j JU-G Avmv cÖkœvejx 1. x 2 + y2 + 3x – 5y + 2 = 0 e„‡Ëi Dci ̄’ (1, 2) we›`y‡Z Aw1⁄4Z ̄úk©‡Ki mgxKiY †KvbwU? [JU 22-23] x – 5y + 3 = 0 5x – 2y – 3 = 0 5x – y – 3 = 0 2x – 5y + 3 = 0 DËi: 5x – y – 3 = 0 e ̈vL ̈v: x 2 + y2 + 3x – 5y + 2 = 0 e„‡Ëi Dci ̄’ (1, 2) we›`y‡Z Aw1⁄4Z ̄úk©‡Ki mgxKiY, 1.x + 2y + 3 2 (x + 1) – 5 2 (y + 2) + 2 = 0  2x + 4y + 3x + 3 – 5y – 10 + 4 = 0  5x – y – 3 = 0 A_ev, Option Test K‡i ïaygvÎ 5x – y – 3 = 0 mgxKiYwU (1, 2) we›`yMvgx| 2. x 2 + y2 – gx = 0 e„Ë Øviv Ave× †ÿ‡Îi †ÿÎdj KZ eM© GKK? [JU 22-23] 1 8 g 2 1 4 g 2 1 2 rg2 rg2 DËi: 1 4 g 2 e ̈vL ̈v: x 2 + y2 – gx = 0 e„‡Ëi †K›`a     g 2  0 I c = 0  r 2 = g 2 4 + 02 – 0 = g 2 4  †ÿÎdj = r 2 =   g 2 4 = 1 4 g 2 eM© GKK 3. r 2 – 2 3rcos – 8rsin + 15 = 0 e„‡Ëi e ̈vmva© KZ GKK? [JU 22-23; CU 22-23] 1 2 3 4 DËi: 2 e ̈vL ̈v: r 2 – 2 3rcos – 8rsin + 15 = 0  x 2 + y2 – 2 3x – 8y + 15 = 0 GLv‡b, g = – 2 3 2 = – 3; f = – 8 2 = – 4; c = 15  e ̈vmva© r = g 2 + f 2 – c = 3 + 16 – 15 = 4 = 2 GKK 4. y = x + 2 mij‡iLvwU x 2 + y2 = 16 e„‡Ë †h R ̈v Drcbœ K‡i Zvi •`N© ̈ KZ? [JU 22-23] 14 2 14 2 2 2 7 DËi: 2 14 e ̈vL ̈v: 4 O(0, 0) A C B x – y + 2 = 0 2 cÖ`Ë e„‡Ëi †K›`a O (0, 0); e ̈vmva© 4 GKK GLb, †K›`a †_‡K †iLvwUi j¤^ `~iZ¡ =     0 – 0 + 2 1 2 + 12 = 2 OC = 2 ; OA = 4; AB = ? wc_v‡Mviv‡mi Dccv` ̈ Abymv‡i- OA2 = OC2 + AC2  AC2 = OA2 – OC2 = 42 – ( 2) 2  AC2 = 16 – 2  AC = 14  R ̈v Gi •`N© ̈, AB = 2  AC = 2 14 GKK