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"

- anonymous

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- 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

- MrNood

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

- anonymous

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

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## More answers

- anonymous

1 sec

- MrNood

ok - sorry - i see my mistake:

- MrNood

|dw:1444259449121:dw|

- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

@amistre64

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

y=mx+b ??

- 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

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

do i find whats m first?

- anonymous

m=7/-4

- amistre64

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

- anonymous

o ok

- 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

- anonymous

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

- 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 :)

- anonymous

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

- 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

- 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

- anonymous

which wolf are u talking about...

- amistre64

wolframalpha.com

- 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

- anonymous

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

- amistre64

that is NOT how you multiply

- amistre64

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

- 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

- amistre64

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

- amistre64

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

- 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

- anonymous

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

- amistre64

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

- 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...

##### 1 Attachment

- 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

- amistre64

|dw:1444264864450:dw|

- anonymous

o ok looks interesting

- 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|

- 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

- 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

- amistre64

|dw:1444265497662:dw|

- 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

- 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

- amistre64

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

- amistre64

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

- 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

- 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

- amistre64

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

- 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?

- amistre64

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

- 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?

- 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 = ??

- amistre64

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

- anonymous

i got y=-1/5

- amistre64

correct

- 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

- anonymous

ok... 1 sec brb

- 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.

- amistre64

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

- amistre64

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

- 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 :)
\[m=\frac{7(-4)\pm\sqrt{16(49)-(16-4(49)-49-13)(-4(49)-13)}}{16-4(49)-49-13}\]

- amistre64

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

- anonymous

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

- 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

- amistre64

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

- anonymous

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

- 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?

- amistre64

|dw:1444270137860:dw|

- anonymous

ok like this |dw:1444266705960:dw|

- 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

- amistre64

good to there?

- amistre64

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

- anonymous

yea good up to there

- 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)

- anonymous

ok

- 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

- 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

- 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 :)

- 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

- 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.

- 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 ..

- 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 :)

- anonymous

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

- 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

- amistre64

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

- amistre64

good luck

- anonymous

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

- anonymous

thank you

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