Title Circle Practice Sheet Solution HSC FRB 25.pdf ✅

e„Ë  Final Revision Batch '25 1 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| mgvavb: (K) GLv‡b, e„‡Ëi civwgwZK mgxKiY, x 2 = 25 – t 2 .... (i) Ges y = t ...... (ii) (i) bs G t = y ewm‡q cvB, x 2 = 25 – y 2  x 2 + y2 = 52  e„ËwUi e ̈vmva© = 5 GKK  e ̈vm = 2  5 = 10 GKK (Ans.) (L) awi, e„ËwUi mgxKiY, x 2 + y2 + 2gx + 2fy + c = 0 ..... (i) e„‡Ëi †K›`a x A‡ÿi Dci n‡j †K‡›`ai †KvwU k~b ̈|  – f = 0  f = 0 Avevi, e„ËwU (0, 0) we›`yMvgx n‡j, c = 0 (2, 3) we›`yMvgx n‡j, 2 2 + 32 + 2 .g.2 + 2.f.3 + c = 0  4 + 9 + 4g = 0 [⸪ f = 0 ; c = 0]  4g = – 13  g = – 13 4 (i) bs G g, f, c Gi gvb ewm‡q wb‡Y©q e„ËwUi mgxKiY, x 2 + y2 – 2  13 4 x + 0 + 0 = 0  4x2 + 4y2 – 26x = 0  2x2 + 2y2 – 13x = 0 (Ans.) (M) GLv‡b, e„ËwUi †K›`a (2, 3) Ges e„ËwU x Aÿ‡K ̄úk© K‡i|  e„‡Ëi e ̈vmva© = |†K‡›`ai †KvwU| = |3| = 3 ZvB e„‡Ëi e ̈vmva© n‡e †KvwUi ciggv‡bi mgvb| myZivs e ̈vmva© = |3| = 3 GKK| (0, 0) we›`yMvgx OC †iLvi Xvj m n‡j mgxKiY, y = mx  mx – y = 0 ...... (i) OC †iLv ̄úk©K n‡j †K›`a (2, 3) †_‡K j¤^ `~iZ¡ n‡e e„ËwUi e ̈vmv‡a©i mgvb|  |2m – 3| m 2 + (– 1) 2 = 3  |2m – 3| = 3 1 + m2  (2m – 3)2 = 9(1 + m2 )  4m2 – 12m + 9 = 9 + 9m2  5m2 + 12m = 0  m (5m + 12) = 0 nq, m = 0 A_ev, 5m + 12 = 0  5m = – 12  m = – 12 5 wKš‘, m = 0, x A‡ÿi Xvj|  OC †iLvi Xvj m = – 12 5 m Gi gvb (i) bs mgxKi‡Y ewm‡q cvB, – 12 5 x – y = 0  – 12x – 5y = 0  12x + 5y = 0 (Ans.)
2  Higher Math 1st Paper Chapter-4 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| mgvavb: (K) GKwU e„‡Ëi †K‡›`ai †cvjvi ̄’vbv1⁄4 (r, ) =     6   4  e„ËwUi Kv‡Z©mxq ̄’vbv1⁄4 (x, y) = rcos, rsin =     6cos  4  6sin  4 =     6 2  6 2 = (3 2  3 2) GLb (3 2  3 2) †K›`a Ges 5 e ̈vmva©wewkó e„‡Ëi mgxKiY, (x – 3 2) 2 + (y – 3 2) 2 = 52  x 2 – 6 2x + 18 + y2 – 6 2y + 18 = 25  x 2 + y2 – 6 2x – 6 2y + 36 – 25 = 0  x 2 + y2 – 6 2x – 6 2y + 11 = 0 (Ans.) (L) GLv‡b e„‡Ëi mgxKiY, x 2 + y2 – 2x + 2y = 2  x 2 + y2 – 2x + 2y – 2 = 0 ....... (i)  e„ËwUi †K›`a (– g, – f) = (1, – 1) Ges e ̈vmva©, r = g 2 + f2 – c = (– 1) 2 + 12 – (– 2) = 1 + 1 + 2 = 4 = 2 GKK e„ËwUi GKwU ̄úk©K, 3x + 4y – 9 = 0 ̄úk©‡Ki j¤^ †iLvi mgxKiY, 4x – 3y + k = 0...... (ii) (ii) bs †iLv (i) bs e„‡Ëi ̄úk©K n‡j †K›`a n‡Z ̄úk©‡Ki j¤^ `~iZ¡ e„‡Ëi e ̈vmv‡a©i mgvb|  |4  1 – 3  (– 1) + k| 4 2 + (– 3) 2 = 2  |k + 7| 5 = 2  |k + 7| = 5  2  k + 7 =  10 nq k + 7 = 10  k = – 7 + 10 = 3 A_ev, k + 7 = – 10  k = – 7 – 10 = – 17 k Gi gvb (iii) bs mgxKi‡Y ewm‡q cvB, k = 3 n‡j mgxKiY, 4x – 3y + 3 = 0 (Ans.) k = – 17 n‡j mgxKiY, 4x – 3y – 17 = 0 (Ans.) (M) GLv‡b, e„‡Ëi mgxKiY, x 2 + y2 – 2x + 2y = 2  x 2 + y2 – 2x + 2y – 2 = 0 ÔLÕ n‡Z cÖvß, e„ËwUi †K›`a (1, – 1) Ges e ̈vmva© = 2 (4, – 3) we›`yMvgx m Xvj wewkó mij‡iLvi mgxKiY, (y + 3) = m(x – 4)  y + 3 = mx – 4m  mx – y – 4m – 3 = 0 ..... (i) (i) bs †iLv e„‡Ëi ̄úk©K n‡j †K›`a (1, – 1) n‡Z ̄úk©‡Ki j¤^ `~iZ¡ e„‡Ëi e ̈vmv‡a©i mgvb|     m + 1  – 4m – 3 m 2 + (– 1) 2 = 2      – 3m – 2 m 2 + 1 2 = 4  9m2 + 12m + 4 m 2 + 1 = 4  9m2 + 12m + 4 = 4m2 + 4  5m2 + 12m = 0  5m   m +  12 5 = 0  m   m +  12 5 = 0  m = 0 A_ev, m + 12 5 = 0  m = – 12 5 m = 0 n‡j ̄úk©‡Ki mgxKiY, y + 3 = 0 (Ans.) m = – 12 5 n‡j ̄úk©‡Ki mgxKiY, – 12 5 x – y – 4     – 12 5 – 3 = 0  – 12 5 x – y + 48 5 – 3 = 0  12x + 5y – 33 = 0 (Ans.) [– 5 Øviv ̧Y K‡i] GLb, (4, – 3) we›`y †_‡K cÖ`Ë e„‡Ëi Dci Aw1⁄4Z ̄úk©‡Ki •`N© ̈ = 4 2 + (– 3) 2 – 2  4 + 2 (– 3) – 2 = 16 + 9 – 8 – 6 – 2 = 25 – 16 = 9 = 3 GKK [ ̄úk©‡Ki •`N© ̈ wbY©‡qi Rb ̈ Aek ̈B e„‡Ëi mgxKiY‡K Av`k© AvKvi x 2 + y2 + 2gx + 2fy + c = 0 G wb‡Z n‡e|] 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|