## anonymous one year ago How do you find the equation of a tangent line and also normal line when given an equation of the circle and a point that is NOT on the circle .. the problem x^2+y^2=13 and point (-4, 7) "find the equation of a tangent line and also normal line when given an equation of the circle and a point that is NOT on the circle"

1. MrNood

|dw:1444258760551:dw| The line that is normal is always a diameter, and passes through the centre. From the equation of th ecircle you know the cntre is (0,0) so you have 2 points on the normal line so can work out its equation. The slope of the tangent is perpendicular to the normal so its slope is -1/m

2. MrNood

because the point is inside the circle there is no tangent through the point...

3. anonymous

no the point is outside the circle ...this is how it look..

4. anonymous

1 sec

5. MrNood

ok - sorry - i see my mistake:

6. MrNood

|dw:1444259449121:dw|

7. anonymous

8. anonymous

yea the lecture give us it and told us to finish it off... stww.. idk what to do..

9. anonymous

@amistre64

10. amistre64

do you know how to determine the equation of a tangent line using derivatives?

11. amistre64

or, we can leave the slope open and solve it after equating the line and the circle equations

12. anonymous

no .. this is geometry ...but if its easier y not show mee

13. amistre64

what would the equation for a line going thru a point having a slope .. be?

14. anonymous

y=mx+b ??

15. amistre64

thats only if we have a special point, the intercept. in general we can use the point slope format y-b = m(x-a) or simply y = m(x-a)+b for a point (a,b) now lets sub this into our circle equation where y is

16. amistre64

x^2+y^2=13 x^2+(m(x-a)+b)^2=13 expand it out

17. amistre64

you can use -4,7 instead of a,b of course

18. anonymous

ok x^2+(m(x+4)+7)^2=13

19. anonymous

ok x^2+(m(x+4)+7)^2=13

20. amistre64

now expand it all out ... im working it on notepad to see if we come to the same conclusions :)

21. anonymous

do i find whats m first?

22. anonymous

m=7/-4

23. amistre64

m is undetermined, it is something we will hopefully find out at the end of the process.

24. anonymous

o ok

25. anonymous

x^2+(m(x+4)+7)^2=13 x^2+(mx+4m+7)^2=13 x^2+m^2x^2+16m^2+49=13

26. anonymous

x^2(1+m^2)+16m^2=-36

27. amistre64

x^2 + (m(x+4)+7)^2 = 13 x^2 + m^2(x+4)^2 +49 +14m(x+4) = 13 x^2 + m^2(x^2+16+8x) +49 +14mx +56 = 13 x^2 + m^2 x^2 +16m^2 +8m^2x +92 +14mx = 0 (m^2+1) x^2 +(8m^2+14m) x +(9 +16m^2) = 0 in order for there to only be one root, the discriminant of this quadratic need to equal zero. (8m^2+14m)^2 -4(m^2+1)(9 +16m^2) = 0 64m^4 +196 m^2 +2(8)(14)m^3 -(4m^2+4)(9 +16m^2) = 0 64m^4 +196 m^2 +2(8)(14)m^3 -(36m^2 +36 +64m^4 +64m^2) = 0 64m^4 +196 m^2 +2(8)(14)m^3 -36m^2 -36 -64m^4 -64m^2 = 0 2(8)(14)m^3 +96 m^2 -36 = 0 i cant be certain that i kept all of that straight :)

28. anonymous

x^2 + m^2(x+4)^2 +49 +14m(x+4) = 13 where did u get the 14m(x+4) part frm?

29. amistre64

gotta rechk that, the wolf says im off i think ... (a+b)^2 = a^2 + b^2 + 2ab if a=m(x+4) and b=7

30. amistre64

and (nm)^2 = n^2 m^2 so im wondering why the wolf says m(x+4)^2 instead of m^2(x+4)^2

31. anonymous

which wolf are u talking about...

32. amistre64

wolframalpha.com

33. amistre64

mx + 4m + 7 mx + 4m + 7 ------------ mmxx +4mmx +7mx 4mmx +16mm + 28m 7mx + 28m +49 ------------------------------------- m^2x^2 +8m^2x +14mx +16m^2 +56m + 49 m^2(x^2 +8x +16) + m(14x+56) + 49

34. anonymous

shouldnt it be mmxx+ 8mm+14 im confuse ...were u adding it?

35. amistre64

that is NOT how you multiply

36. amistre64

i should prolly ask how you ahve been solving these before i came in to muddy the waters

37. anonymous

(mx+4m+7)(mx+4m+7) mxmx+mx4m+7mx+4mmx+16mm+28m+7mx+28m+49 m^2x^2+4m^2x+7mx+4m^2x+16m^2+28m+7mx+28m+49 m^2x^2+8m^2x+14mx+16m^2+56m+49

38. amistre64

|dw:1444263458406:dw| what is the value of n?

39. amistre64

n^2 = 4^2+7^2-13 = 52 n = sqrt(52) right?

40. amistre64

now we know the radius of a circle centered at -4,7 and can equate 2 circles to find where they meet (x+4)^2 + (y-7)^2 - 52 = x^2 + y^2 -13

41. anonymous

ok i will be back in the next 20 second ...

42. amistre64

i got it figured out :) ill be back as well ... food is on the table

43. anonymous

but how can the circle be center at (-4,7) when thats the point the tangent pass through and that point is not on the circle ...this was what the lecture give... and then tell us figure our the rest...

44. amistre64

you can center a circle at any point you wish ... my solution is to create a circle at the given point, with a radius that reaches to the point of tangency ... then solve the 2 equations to see where their solution set is

45. amistre64

|dw:1444264864450:dw|

46. anonymous

o ok looks interesting

47. amistre64

so, i would determine the distance from the given point to the origin (the center of our given circle) as it forms the hypotenuse of a right triangle with the radius of the given circle as a leg. the radius of our unknown circle is the missing leg |dw:1444264994717:dw|

48. amistre64

n = sqrt(52) so we our circle of radius sqrt(52) centered at -4,7 is (x+4)^2 + (y-7)^2 = 52 and equate this with the given: x^2 + y^2 = 13 (x+4)^2 + (y-7)^2 +13 = 52 +x^2 + y^2 x^2+8x+16 + y^2-14y+49 +13 = 52 +x^2 + y^2 8x+16 -14y+49 +13 = 52 we have reduced the solution to a linear equation; solve for y 8x -14y= 52 -16 -13 -49 8x -14y= -26 4x -7y= -13 i wonder of that is a coincidence ... y = (4x+13)/7

49. amistre64

this is actually the line the passes between the points of tangency ... it makes sense since 2 points create a line to start with and the solution set is 2 points. sub this in to find x x^2 + y^2 = 13 x^2 + [(4x+13)/7]^2 = 13

50. amistre64

|dw:1444265497662:dw|

51. anonymous

im not sure if i follow all this ..but x^2+(4x+13)^2/49=13 49x^2+16x^2+104x+169=637 65x^2+104x-468=0 5x^2+8x-13=0 5x^2-5x+13x-13=0 5x(x-1)+13(x-1)=0 5x+13=0 x=-13/5 and x-1=0 x=1

52. amistre64

d^2 = (a^2+b^2), for some point (a,b) n^2 = a^2+b^2 - k, for some given r^2=k (x-a)^2 + (y-b)^2 +k = n^2 + x^2 + y^2 x^2 -2ax + a^2 + y^2 -2by +y^2 +k = n^2 + x^2 + y^2 -2ax + a^2 -2by +b^2 +k = n^2 -2ax + a^2 +b^2 +k - n^2 = 2by [-2ax +a^2 +b^2 +k -(a^2+b^2 - k)]/2b = y [-2ax +2k]/2b = y [-ax +k]/b = y

53. amistre64

x=-18/5 and x=2 i think you might have made a few errors

54. amistre64

so it does generalize into a line solution of y = (-ax+k)/b y = (4x+13)/7 in this case

55. amistre64

x^2 + (4x+13)^2/49 = 13 49x^2 + 16x^2 +169 +8(13)x = 13(49) 65x^2 +104x -468 = 0 13 (x-2) (5x+18) = 0

56. amistre64

using y = (4x+13)/7 with those x values we can solve for y ... and all this does is gives us the tangency points which we can then find the slopes (m1,m2) for the lines y = m1(x+4) + 7 and y = m2(x+4) + 7

57. amistre64

or as noted, y' = -x1/y1 will give us the slopes as well

58. anonymous

omg all this working to 1 question...i wonder how much marks if this question come as a test question in the next 2 weeks..would it go for...stw... hope this dont come...any how what does m1=? did we find that?

59. amistre64

you have to determine your points of tangency ... we found the x values (-18/5 and 2), what are the y values ?

60. anonymous

y=m1(x+4)+7 when x=-18/5 y=m1(-18/5+4)+7 y=m1(2/5)+7 am i going right?

61. amistre64

no, the solution set is on the line: y = (-ax+k)/b y = (4x+13)/7 when x=2 ... y = 21/7 = 3 when x=-18/5 ... y = (4(-18)/5+13)/7 = ??

62. amistre64

the slope of that tangent at 2,3 is: -2/3 , the normal slope is 3/2

63. anonymous

i got y=-1/5

64. amistre64

correct

65. amistre64

y = m(x-a)+b x^2 + m^2 (x^2 -2ax + a^2) + b^2 +2mb(x-a) = k x^2 + m^2 x^2 -2a m^2 x + (am)^2 + b^2 +2mb x- 2mba -k = 0 (m^2+1) x^2 +2m(b-am) x + (am)^2 + b^2 - 2mba -k = 0 this has one root value if the discriminant is equal to 0 soo (2m(b-am))^2 -4((m^2+1))((am)^2 + b^2 - 2mba -k) = 0 when we let a=-4, and b=7 (2m(7-(-4)m))^2 -4((m^2+1))(((-4)m)^2 + 7^2 - 2(7)(-4)m -13) = 0 the wolf simplifies that to m=-2/3 and m=-18 for our slopes

66. anonymous

ok... 1 sec brb

67. amistre64

(m(x-a)+b)^2 x^2 + m^2 (x^2 -2ax + a^2) + b^2 +2mb(x-a) = k x^2 + m^2 x^2 -2a m^2 x + (am)^2 + b^2 +2mb x- 2mba -k = 0 (m^2+1) x^2 +2m(b-am) x + (am)^2 + b^2 - 2mba -k = 0 (2m(b-am))^2 -4((m^2+1))((am)^2 + b^2 - 2mba -k) = 0 (2m(7-(-4)m))^2 -4((m^2+1))(((-4)m)^2 + 7^2 - 2(7)(-4)m -13) = 0 4m^2 (b-am)^2 -(4m^2+4)((am)^2 + b^2 - 2mba -k) = 0 (am)^2 + b^2 - 2mba -k 4m^2 + 4 ------------------------ 4a^2m^4 +4ab^2 m^2 -8abm^3 -4km^2 +4a^2 m^2 +4ab^2 -8abm -4k 4(a^2) m^4 -4(2ab) m^3 +4(a^2+ab^2-k) m^2 -4(2ab) m +4(ab^2-k) divide off the 4 (a^2) m^4 -(2ab) m^3 +(a^2+ab^2-k) m^2 -(2ab) m +(ab^2-k) ---------------- equals 0 yeah, i dont think there is anyway that you would want to work the formula for finding a roots of a quartic equation.

68. amistre64

might be some calc errors in there cause the wolf says we obtain a quadratic with your a,b

69. amistre64

oh, thats just the 4ac part ... negate and add in the b^2 might help

70. amistre64

4m^2 (b-am)^2 4m^2 (b^2 +(am)^2 -2abm) 4(a^2) m^4 -4(a^2) m^4 -4(2ab) m^3 +4(2ab) m^3 4 (b^2) m^2 -4(a^2+ab^2-k) m^2 +4(2ab) m -4(ab^2-k) -(a^2+ab^2-b^2-k) m^2 +(2ab) m -(ab^2-k) $m=\frac{ab\pm\sqrt{a^2b^2-(a^2+ab^2-b^2-k)(ab^2-k)}}{a^2+ab^2-b^2-k}$ not too sure how useful this would be ... doesnt look simple to memorize :) $m=\frac{7(-4)\pm\sqrt{16(49)-(16-4(49)-49-13)(-4(49)-13)}}{16-4(49)-49-13}$

71. amistre64

id stick with that: y = (-ax+k)/b and plugging that into the circle

72. anonymous

ok y do we have to do this part ...what are we finding for ? frm my last post to this post...

73. amistre64

we are finding the x and y values for the points of tangency to the given circle so that we can define sloped for the lines of tangency/normality

74. amistre64

if you know of some other way ... feel free

75. anonymous

no urs is ok... ill figure it out... so whats the next step... and i really appreciate ur help

76. amistre64

thru generalization, i determined that the solution points will form the line: y = (-ax+k)/b this is the line that connects the 2 points of tangency; does this make sense?

77. amistre64

|dw:1444270137860:dw|

78. anonymous

ok like this |dw:1444266705960:dw|

79. amistre64

yes, those are the line equations we need to determine, but we need to find the points of tangency on the circle using the point -4,7 and a circle equal to 13 y = (4x+13)/7 will help us get our points of tangency x^2 + y^2 = 13, sub in for y x^2 + [(4x+13)/7]^2 = 13 we determined that x=2 and -18/5 y = (4(2)+13)/7 = 21/7 = 3; (2,3) is one point y = (4(-18/5)+13)/7 = -1/5; (-18/5, -1/5) is the other point

80. amistre64

good to there?

81. amistre64

works coming early in the morning so i have to wrap this up soon

82. anonymous

yea good up to there

83. amistre64

since we know our point of tangency (x,y) our slope of tangency is just -x/y (and slope of normalcy is y/x)

84. anonymous

ok

85. amistre64

we then construct our lines using our tangency points and the slopes required. y-b = m(x-a) for tangent points (a,b) and slope m

86. amistre64

for the point (2,3) , tangent slope is -2/3 and normal slope is 3/2 lines are: tan y = -2/3 (x-2)+3 norm y = 3/2 (x-2) +3

87. amistre64

it just occured to me that the normal line is going thru the origin anyways so y = 3/2 x is what it simplifies to :)

88. anonymous

ok and the point (-18/5, -1/5).. tangent slope is -18 and normal slope is 1/18 lines are tan y=-18(x+18/5)-1/5 normal y=1/18(x+18/5)-1/5

89. amistre64

yep, norm y can be simplified as y = 1/18 x the constant zeros out becuase the y intercept is the origin for the normal line.

90. anonymous

ok cool... one question... did u wrote all this out on paper? i want to see the consistency of each step like without my part... can u attach it ..

91. amistre64

i did not, i wrote it out in the post and whats left in my head is bound to flitter away int oblivion by tomorrow :)

92. anonymous

ok thanks ..i will go through back all the post to make sure i understand everything

93. amistre64

my idea is only one method .... im sure there is a simpler approach. but what i did was: i found that the solution points generalize to the line: y = (-ax+k)/b sub that into the circle to determine x values; which in turn can be used to define y

94. amistre64

use the points of tangency to define slopes and lines as needed

95. amistre64

good luck

96. anonymous

i think my problem here is ..i dont fully understand this tangency thing ...i will watch some videos on it..

97. anonymous

thank you