Properties of Conics

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1 Discipline Courses-I Semester-I Pper: Clculus-I Lesson: Properties of Conics Lesson Developer: Hrindri College/Deprtment: Deprtment of Mthemtics, Deshbndu College, University of Delhi Institute of Lifelong Lerning, University of Delhi pg.

2 Tble of Contents: Chpter : Properties of Conics : Lerning Outcomes : Introduction 3: Eccentricity o 3.: Eccentricity of Ellipse o 3.: Eccentricity of Hyperbol 4: Discriminnt Exercise References. Lerning Outcomes: After you hve red this chpter, you should be ble to Define the eccentricity, Define nd find the eccentricity of ellipse, Define nd find the eccentricity of hyperbol, Identify the conic with the help on discriminnt. Institute of Lifelong Lerning, University of Delhi pg.

3 "I know not wht I pper to the world, but to myself I seem to hve been only like boy plying on the seshore, nd diverting myself in now nd then finding smoother pebble or prettier shell, whilest the gret ocen of truth ly ll undiscovered before me." SIR ISAAC NEWTON "The method of fluxions the clculus is the generl key by help whereof the modern mthemticins unlock the secrets of Geometry nd consequently of Nture" BISHOP BERKELY. Introduction: Modern mthemtics begn with two gret dvncesnlytic geometry nd the clculus. Anlytic geometry took definite form in the yer 637 while the clculus took the definite shpe in 666. Rene Descrtes, the gret French mthemticin of the seventeenth century, solved the problem of describing the position of point in plne. His method ws development of the older ide of ltitude nd longitude. In honour of Descrtes, the system used for describing the position of point in plne is lso known s the Crtesin system. When he introduced nlytic geometry he ws interested in it s n id in the solution of eqution. It soon developed, however, into method of solving problem in plne nd solid geometry lgebriclly, gretly influenced the study of both geometry nd lgebr. Anlytic geometry is divided into two brnches: plne nlytic geometry, deling with points, lines nd curves which re restricted to plne; nd solid nlytic geometry, deling with points, lines, plnes, curves nd surfces in the threedimensionl spce but in this chpter we study plne nlytic geometry. On the other hnd, both Newton nd Leibnitz, shre the credit for inventing "Clculus" independently in the seventeenth century. The former used physicl pproch while the ltter used geometricl pproch. Further Newton used the term "rte of chnge" in his second lw of motion. Thus the clculus, sometimes, my be defined s mthemtics of motion nd chnge. Conic section is curve obtined s the intersection of doublenpped right circulr conic with plne. Conics re brodly clssified s degenerte conic which includes point, line nd pir of intersecting lines or nondegenerte conic which includes circle, prbol, ellipse nd hyperbol. The conics re the pth trvelled by plnets, stellites nd other bodies whose motion re driven by inverse squre forces. Institute of Lifelong Lerning, University of Delhi pg. 3

4 In this chpter we will discuss focusdirectrix chrcteriztion of conic section which introduces the concept of eccentricity nd is defined for degenerte conics. The eccentricity of conic section in mesure of how fr it divites from being circulr. Rene Descrtes pplied his newly discovered nlytic geometry to the study of conics which further reduces the geometricl problem of conics to problem in lgebr. It ws lso observed tht every generl eqution of second degree represents conic under certin conditions nd for tht discriminnt test is lso estblished. Conics hs wide rnge of ppliction, for exmple, every prtil differentil eqution is clssified s prbolic, elliptic or hyperbolic. The behviour nd theory of these different types of prtil differentil eqution is strikingly different representtive exmples is tht the Poisson eqution is elliptic, the het eqution is prbolic nd the wve eqution is hyperbolic. Furthermore, conics hve mny prcticl ppliction lso. The prbol hs vrious technicl ppliction. Fmilir exmples re the prbolic reflector, the prbolic rch nd the prbolic suspension of cbles. The property used in such cses in clled the reflective property of the prbol. It is observed tht sound wves follow elliptic pths nd the reflective property of ellipse is used in constructing whispering glleries. Wter pipes re sometimes designed with ellipticl cross-section to llow for expnsion when the wter freezes. Also, hyperbolic pths rise in Einstein's theory of reltivity nd form the bsis for the LORAN (Long rnge nvigtion) rdio nvigtion system. The chpter ends with the sttement nd proof of the reflective properties of the prbol, ellipse nd hyperbol. We observe tht conic is just section or slice through cone. Institute of Lifelong Lerning, University of Delhi pg. 4

5 We notice tht the conics such s circle, prbol, ellipse nd hyperbol re defined only using stright line clled the directrix nd point (clled the focus). Now mesure the distnce from the focus on the curve nd the distnce perpendiculrly from the directrix to tht point (see figure ). We observe tht these two distnces will lwys be in some rtio. Bsed on these two distnces we ssocite number clled eccentricity with ech conic section. 3. Eccentricity: Eccentricity, denoted by e, of curve is defined s e Dis tn ce from the focus to point on the curve Perpendiculr dis tn ce from the directrix to tht point PF PM () Institute of Lifelong Lerning, University of Delhi pg. 5

6 So conic section my be re-defined s the locus of ll points whose distnce to the focus is equl to the eccentricity times the distnce to the directrix. 3.. Eccentricity of Ellipse: The eccentricity of the ellipse c b defined s e where > b. 3.. Eccentricity of Hyperbol: The eccentricity of the hyperbol x y is defined s b Institute of Lifelong Lerning, University of Delhi pg. 6 x y b () is c b e (3) Vlue Addition: Remrks () We conclude, in cse of prbol, e =. (b) Note tht for n ellipse we hve 0e becuse the foci re closer together thn the vertices. (c) For hyperbol, we conclude tht e>, becuse the foci re frther prt thn the vertices. Further, if the eccentricity is smller, The curve is more curved nd if it is bigger, then the curve is less curved. (d) We observe tht circle of rdius my be regrded s the limiting cse of n ellipse whose mjor xis is nd whose eccentricity tends to zero. So the eccentricity shows how "un-circulr" the curve is. The bigger the eccentricity, the less curved it is (see fig. 3). (e) The ltus rectum is chord of the ellipse pssing through the focus nd perpendiculr to the xis. Its length b is (where is the semi mjor xis nd b is the semi minor xis)

7 (f) (g) By definition, MF M is the ltus rectum. Then MFM MF Let MF = y, thus the coordintes of M is (-e, y) Since M lies on the ellipse, therefore we hve e y. b i. e., 4 b b y b e by using e Therefore y b b. Hence ltus rectum The ltus rectum is chord of the hyperbol pssing through the focus nd perpendiculr to the focl xis. Its length is b (for proof see remrk (e)). We observe tht the directrices of n ellipse re given by x which e corresponds to the foci (±c, 0). Exmple : Find the eccentricity, coordintes of the foci, nd the length of the ltus rectum of the ellipse 4x + 9y =. Also find its directrices. Solution: The eqution of the ellipse cn be re-written s x y. 4 9 This implies tht, b (by compring with eqution (7)). 4 9 From eqution (6), c b e 3 5 Thus, coordintes of the foci re e,0, i. e.,, 0 6 Institute of Lifelong Lerning, University of Delhi pg. 7

8 Also, the length of the ltus rectum b Hence, the eqution of the directrices re x 3 e 5 Exmple : Find the length of the xes nd of the ltus rectum of the ellipse 4x + 3y = 4. Solution. Rewriting the given eqution, we hve x y 6 8 By compring with eqution (7), we obtin = 6, b = 8 Thus in this cse the x-xis lies long the minor xis nd yxis long the mjor xis nd therefore the lengths of the xes re 8 nd 6. Also, the length of the ltus rectum b 6 3. Exmple 3: Find the centre nd eccentricity of the ellipse x + 3y - 4x + 5y + 4 = 0 Solution: Re-writing the given eqution, we obtin So tht, x x 3y y y x Institute of Lifelong Lerning, University of Delhi pg. 8

9 Let X = x, Y = y + 5. Then the given eqution becomes 6 X Y 4 36 Thus, the given eqution represents n ellipse whose centre is t 5, nd ; b Now, b 4 36 e nd hence 3 4 e. 3 Exmple 4: Find the eqution of the hyperbol centred t the origin of the xyplne tht hs focus t (,0) nd the line x s the corresponding directrix. Solution: By definition, the focus is (c, 0) = (, 0) so tht c =. Now eqution of the directrix is given by x, i. e., e. e c 4 Also, e, e e i.e., e = 4 nd hence e =. Let P(x, y) be ny point on the hyperbol. We know tht PF = e.pm (by using eqution (5)) i. e., x y 0 x Institute of Lifelong Lerning, University of Delhi pg. 9

10 i. e., x y 4 x squring both sides i e x y.., 3 3 x y ie.., 3 which is the required eqution of the hyperbol. Exmple 5: Find the eqution of the ellipse centred t the origin of the xyplne tht hs focus t (0, ±3) nd 0.5 s its eccentricity. Solution: The focus F (0, ±c) is (0, ±3) so tht c = 3. Now, e c, i.e., so tht = 6. Also, c = b, i.e., 9 = 36 b which gives b = 7. Thus, the eqution of the ellipse is given by x y Exmple 6: Find the centre eccentricity, foci nd direction of the hyperbol 9x - 6y = 44. Solution: The given hyperbol cn be re-written s x y. 6 9 This implies tht = 6, b = 9. Then, b e (by using eqution (3)). 4 4 Also, centre is t (0, 0). Now, c = e = Institute of Lifelong Lerning, University of Delhi pg. 0

11 Thus, foci t (±5, 0). Hence the eqution of the directrices re given by 4 6 x. e Exmple 7: A hyperbol of eccentricity 3/ hs one focus t (, -3). the corresponding directrix is the line y =. Find n eqution for the hyperbol. Solution: Let P(x, y) be ny point on locus of the hyperbol. Then, by using eqution (), we hve PF 3 i. e., x y 3 y e PM. 9 i. e., x y 3 y squring both the sides 4 i. e., 4 x y 6y 9 9 y 4y 4 i. e., 4 x 5 y y i. e., 4 x 5 y y6 x ie.., which is the required eqution of the hyperbol. Vlue Addition: Remember Consider second degree eqution of the form Ax + Cy +Dx+Ey+F = (4) By using the vrious fcts bout conics we conclude tht eqution (4) represents () circle if A=C 0 (specil cses: the grph is point or there in no grph). (b) prbol if eqution (4) is qudrtic (second degree) in one vrible nd liner in the other. (c) n ellipse if A nd C re both positive nd n imginry ellipse if both A nd C re negtive. (specil cses: circles, single point or no grph Institute of Lifelong Lerning, University of Delhi pg.

12 (d) (e) (f) t ll) hyperbol if A nd C hve opposite signs (specil cses: pir of intersecting lines). stright line if A=0 nd C=0 nd t lst one of D nd E is different from zero. One or two stright lines if L.H.S. of eqution (4) cn be fctored into the product of two liner fctors. Vlue Addition: Note Now the most generl form of second degree eqution is given by Ax +Bxy+Cy +Dx+Ey+F= (5) where A,B nd C re not ll zero. () If B=0, then eqution (5) reduces to eqution (4). (b) In eqution (4) we notice tht B=0 s the xes of the cones re prllel to the coordinte xes (counter clockwise rottion). Theorem : Let the eqution Ax + Bxy + Cy + Dx + Ey + F = 0 be such tht B 0 nd if nd xy coordinte system is obtined by rotting the xyxes through n ngle stisfying tn eqution 5) will hve the form B A C 0. A x C y D x E y F, then, in xy coordinte, Proof: We will rotte the coordinte xes through n ngle in order to eliminte the xy term, i.e., to find the condition on so tht B = 0. From fig. 5, we hve x = ON=OP cos ( ) = OP cos cos - OP sin sin Also, y = PN=OP sin ( ) = OP sin cos + OP cos sin But OP cos = ON = x nd OP sin = N P = Y Thus the equtions for rotting coordintes xes re given by x xcos y sin y xsin y cos 6 Institute of Lifelong Lerning, University of Delhi pg.

13 Substituting x nd y from the eqution (6) in the eqution (5) we obtin A x + B x y + C y +D x +E y + F = (7) where A Acos Bsin cos sin cos sin B B C A C Asin Bsin cos C cos D D cos E sin E Dsin E cos F F 8 Now choose in such wy tht the coefficient of x y, i.e, B in eqution (7) my vnish. Thus, B cos + (C-A) sin = 0 (by using eqution(8)) i.e, tn = B A C A C or cot (9) B which is the required condition on so tht the eqution (7) becomes Ax C y Dx Ey F 0 which completes the proof. Vlue Addition: Remrks For ny rottion of xes, we hve (i) B - 4AC = B' 4A'C' (ii) A + C = A' + C' This shows tht the bove expressions does not lter under the rottion of xes. The next result shows tht without rotting the coordinte xis (i.e., without eliminting the xyterm from eqution (7)), it is possible to clssify wht type of conic section the eqution (7) represents. Ax + Bxy + Cy + Dx + Ey + F = 0 be ny second- Theorem : Let degree eqution. () (b) It B 4AC < 0, then the eqution represents n ellipse, circle, point or else hs no grph It B 4AC > 0, then the eqution represents hyperbol or pir of interesting lines. Institute of Lifelong Lerning, University of Delhi pg. 3

14 (c) It B 4AC = 0, then the eqution represents prbol, line, pir of prllel lines or else hs no grph. Proof: By using theorem, we obtin 0 7 A x B x y C y D x E y F where A, B, C, D, E, nd F re given by eqution (8) nd choose is given by eqution (9). Then eqution (7) reduces to (0) A x C y D x E y F For ny rottion of xes, we hve sin 0. B AC B A C A C ce B i e B AC AC.., 4 4 () If B 4AC < 0, then by eqution (), we hve AC 0. We first ssume tht A 0 nd C 0. Now we divide eqution (0) by AC so tht it cn be written in the from D E F x x y y C A A C AC D E D E F i. e., x y C A A C 4A C 4AC AC ie.., D E x y A C C A 4A C where CD AE 4ACF x h y k ie.., C A Institute of Lifelong Lerning, University of Delhi pg. 4

15 D E where h, k. A C There re three cses to discuss. (i) Let >0. Then from the eqution (), we conclude tht the grph is either circle if A C or n ellipse if A C (ii) Let = 0. Then, the eqution () is stisfied only by x h nd y k implies tht the grph in this cse is the single point (h, k) which (iii) Let < 0. In this cse there is no grph, since the left side of eqution () is nonnegtive for ll x nd y. Similr rguments holds true for the cse when A 0 nd C 0 so tht AC 0. (b) If B 4AC > 0, then by eqution () we hve AC 0 which implies tht either A 0 nd C 0 or A 0 nd C 0. We ssume tht A 0 nd C 0. Agin there re three cses to discuss. (i) Let > 0. Then from eqution () we deduce tht the grph is hyperbol of the form y k x h b where A nd b C. (ii) Let < 0. Institute of Lifelong Lerning, University of Delhi pg. 5

16 Then the eqution () represents grph of the hyperbol of the from where x h y k (iii) Let = 0. C nd b A Then eqution () reduces to A xh C y k 0. b which is seprble into liner fctors becuse A nd C hve opposite signs. Hence, eqution () in this cse represents two stright lines intersecting in the point (h, k), i.e, D E, A C Likewise, the cse for which A 0 nd C 0 cn be discussed. (c) If B 4AC = 0. then by eqution () we obtin AC 0. which implies tht either A 0 or C 0 or both re zero. (i) Let A 0. Then eqution (0) reduces to 0. C y D x E y F D E F i e y x y C C C.., 0 which represents grph of prbol provided D 0 the xxis xis prllel to If D 0, then eqution (0) represents pir of prllel lines or hs no grph. Institute of Lifelong Lerning, University of Delhi pg. 6

17 (ii) Let C 0. Then eqution (0) becomes 0 A x D x E y F D E F i e x x y A A A.., 0 which represents grph of the prbol provided E 0 hving its xis prllel to the yxis. If E 0, then eqution () represents pir of prllel lines or hs no grph. (iii) Let A 0 nd C 0. Then eqution (0) represents line if either D 0 or E Discriminnt: The term B 4AC in the theorem, is clled the discriminnt of the eqution (5). Vlue Addition: Note Sometimes theorem, is lso known s discriminnt test. Exmple 8: Identify the grph of the conic x xy y x y (3) Solution. We hve A, B 3, C 3, D 3, E, F 0 so tht B AC Thus, the given eqution represents prbol, line, pir of prllel lines or hs no grph. Now, 3 tn B 3 AC 3. Then so tht. 3 6 Institute of Lifelong Lerning, University of Delhi pg. 7

18 Further x xcos ysin 6 6 3x y x y 3 nd y xsin ycos 6 6 On substituting x' nd y' in the eqution (3), we obtin 3 x y 3 y x y 3 x y 3 3 x 3 3 y x y 3 3 x 0 i.e., 3x 3 y 3 xy 3x 3xy xy 3y x y xy x y x y i.e., y x x x y y xy 3xy xy x i.e., y x i.e., 4y x, i. e., y x 4 x 8 so tht. 8 Thus the given eqution represents prbol. Exmple 9: Trce the conic 3x + xy + 3y = (4) Solution. By compring eqution (4) with eqution (5) we obtin A = 3, B =, C = 3 nd F = -9 Consider B 4AC = () = 4 36 = 3 < 0. Institute of Lifelong Lerning, University of Delhi pg. 8

19 Hence the curve (4) is n ellipse, point, circle or else hs no grph Now, tn B AC 33 Then,. 4 The rottion eqution re given by x x xcos ysin ; y xsin ycos x y x y i. e., x ; y sin (5) Substituting eqution (5) in the eqution (4), we get x y x y x y x y which on simplifiction gives 8x 4y 38 x y which is the required eqution of ellipse. Exmple 0: Identify the conic nd lso trce its grph. x + xy + y - = (6) Solution: A =, B =, C =, D = E = 0, F = - Now, B 4AC = 4 = -3 < 0 Therefore (6) represents n ellipse. Also, B tn i. e., i. e., AC 4 Institute of Lifelong Lerning, University of Delhi pg. 9

20 Then, x y x xcos ysin x y y xsin ycos 4 4 On substituting eqution (7) in eqution (6), we obtin x y x y x y x y 0 i. e., x xy y x xy x y y x xy y 0 x x x y y y ie.., x y ie., which represent n ellipse. 3 Exmple : Find the sine nd cosine of n ngle through which the coordinte xes cn be rotted to eliminte the cross product term from the eqution. Also identify the conic. Institute of Lifelong Lerning, University of Delhi pg. 0

21 4x 4xy y 8 5x 6 5y 0 8 Solution: Here A 4, B 4, C, D 8 5, E 6 5, F 0. Consider B 4AC = (-4) 4 4 = 6 6 = 0 Therefore, eqution (8) represents prbol. Now, B 4 4 tn AC 4 3 (by using theorem.3) then 3 cos so tht 5 cos cos cos 5 nd sin cos sin. 5 Also, x y x y x, y Therefore the given eqution becomes x 4xy 4y x xy yx y 4x y 4xy x y x y i.e., y x x x xy xy xy y y x 6y 3x 6y 0 Institute of Lifelong Lerning, University of Delhi pg.

22 i.e., 3 y x which represents prbol. 5 Now we will discuss the reflective properties of prbol, ellipse & hyperbol which useful in different types of pplictions. Theorem 3: The tngent line t point P on Prbol mkes equl ngles with the line through P prllel to the xis of symmetry nd the line through P nd the focus. Proof: Let us consider point P(t, t) on the prbol. Further, suppose tht PT is the tngent to the prbol y = 4x (9) nd F (, 0) is the focus in the figure 9. Also, L is the directrix nd PM is the line prllel to the xis of symmetry We first find the eqution of the tngent PT to the given prbol. For this, consider ny two point P(t, t) nd Q(x, y ) on the given prbol. Then eqution of the chord PQ in given by t y y t x t (0) t x Since Q lies on (9), therefore y 4x By substituting () in (0), we obtin t y y t x t y t 4 4 i. e., y t x t t y Now, the chord PQ will be tngent to the prbol (9) if Q coincides with P, i.e., if t = x nd t = y. Institute of Lifelong Lerning, University of Delhi pg.

23 4 t t Then, y t x t i. e., ty x t which is the required eqution of the tngent PT to the given prbol. Then slope of PT is t t The slopes of PM nd FP re 0 nd t respectively Let TPM =, FPT =. 0 m m Then tn t m m.0 t t nd tn t t t t. t t t Thus, tn = tn, i.e = nd hence the proof is complete. Theorem 4: A line tngent to n ellipse t point P mkes equl ngles with the lines joining P to the foci. Proof: Let the eqution of n ellipse be x y. b Tke point P(x, y ) on it. Suppose tht the tngent nd norml t P meet the xxis (in this cse the mjor xis) in the points M nd N respectively. By using the similr rguments s used in the theorem 3, the eqution of the tngent PT to the given ellipse is s follows. Institute of Lifelong Lerning, University of Delhi pg. 3

24 xx yy b Now, slope of the tngent PT x bx y y b so tht the slope of the norml PN = y bx. Thus, eqution of the norml PN t (x, y ) is given by y bx y y x x ie.., y y x x y x b Therefore, norml cuts the xxis (put y = 0) when b x x e x. This implies tht ON = e x Thus F N = F O + ON = e + e x = e ( + ex ) = ef P nd F N = F O ON = e - e x = e ( ex ) = ef P FP FN Hence, F P F N which concludes tht PN is the bisector of the ngle F PF. F PT F PN NPF F PM Now, Thus, FPT F PM, i. e., which completes the proof. Institute of Lifelong Lerning, University of Delhi pg. 4

25 Theorem 5: A line tngent to hyperbol t point P mkes equl ngles with the lines joining P to the foci. Proof: eqution x Let the eqution of the hyperbol be represented by the y. b Consider ny point P(x, y ) on the hyperbol nd let the tngent PT t P meets the xxis (i.e., focl xis) t the point N. Then, the eqution of the norml t P to the hyperbol is x x y y x y b Now it meets the focl xis (i.e., y = 0) t the point M Thus, OM = e OQ x b x e x so tht F M = OM OF, = e OQ e. = e(ex ) = ef P. By using the sme rguments, we obtin F M = ef P FM FP This implies tht. F M F P i.e., PM is the bisector (externl) of the F PF. Since the tngent is norml to the bisector, therefore it bisects the F PF internlly which completes the proof of the theorem. Institute of Lifelong Lerning, University of Delhi pg. 5

26 Exercise:. Find the eccentricity nd equtions of the directrices of the ellipse x + 9y + 0x + 6 = The line x 0 is one of the directrices of n ellipse, vertex is t 4 (3, ), nd the centre is t (, ). Find the eqution of the ellipse. Also trce the grph. 3. Find the eqution of the hyperbol contining the point (5, ) nd hving the symptote xy = 0. Also find the eccentricity. 4. Find the eqution of the hyperbol hving focus t the point (5, 0) nd vertex t the point (3, 0). 5. Find the locus of point which moves so tht its distnce from the 8 point (0, 0) is fivehlves of its distnce from the line x 0. 5 In ech of the exercises 60, obtin trnsformed eqution of the given curve free from term x'y' nd then identify the conic. Also trce the grph. 6. 4x + 4xy + y 48x 44y + 4 = x + 6xy + y + 4x + 4y 3 = 0 8. x + xy + y x 4y 7 = x - 4xy + 9y + 30x + 90y = x 4xy + y 9x = 0 Solutions:. 59 e ; x x + 5y + 36x 50y 64 = 0 3. x 4y = 9 ; e 5 Institute of Lifelong Lerning, University of Delhi pg. 6

27 4. 6x 9y = x 4y = x y x y hyperbol , x y x y ellipse , x y x y ellipse , y 6x y 0, prbol x y x y ellipse , Summry: In this chpter, we hve emphsized on the followings definition of the eccentricity, how to find the eccentricity of ellipse, how to find the eccentricity of hyperbol, how to identify the conic with the help on discriminnt. References:. Anton, H., Bivens I., nd Dvis, S., "Clculus", Wiley Indi, 0. Cell, J.W., "Anlytic geometry", Toppon compny Ltd., Tokyo, Jpn, Fuller, G., "Anlytic geometry", AddisionWesley Publishing compny. Inc., Holmes, C.T., "Clculus nd Anlytic geometry", Mc grwhill Book Compny, Inc Tylor, A.F., nd Hlberg Jr. Chrles, J.A. "Clculus with nlytic geometry" Prentice Hll Interntionl Inc, Englewood cliffs, N.J., Thoms, Jr. G.B. nd Finney, R.L. "Clculus nd Anlytic Geometry", Person Eduction Inc., 0. Institute of Lifelong Lerning, University of Delhi pg. 7

I. Equations of a Circle a. At the origin center= r= b. Standard from: center= r=

I. Equations of a Circle a. At the origin center= r= b. Standard from: center= r= 11.: Circle & Ellipse I cn Write the eqution of circle given specific informtion Grph circle in coordinte plne. Grph n ellipse nd determine ll criticl informtion. Write the eqution of n ellipse from rel