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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"
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"

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MrNood
 one year ago
Best ResponseYou've already chosen the best response.0dw: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
 one year ago
Best ResponseYou've already chosen the best response.0because the point is inside the circle there is no tangent through the point...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no the point is outside the circle ...this is how it look..

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0ok  sorry  i see my mistake:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea the lecture give us it and told us to finish it off... stww.. idk what to do..

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2do you know how to determine the equation of a tangent line using derivatives?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2or, we can leave the slope open and solve it after equating the line and the circle equations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no .. this is geometry ...but if its easier y not show mee

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2what would the equation for a line going thru a point having a slope .. be?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2thats only if we have a special point, the intercept. in general we can use the point slope format yb = m(xa) or simply y = m(xa)+b for a point (a,b) now lets sub this into our circle equation where y is

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2x^2+y^2=13 x^2+(m(xa)+b)^2=13 expand it out

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2you can use 4,7 instead of a,b of course

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok x^2+(m(x+4)+7)^2=13

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok x^2+(m(x+4)+7)^2=13

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2now expand it all out ... im working it on notepad to see if we come to the same conclusions :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do i find whats m first?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2m is undetermined, it is something we will hopefully find out at the end of the process.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x^2+(m(x+4)+7)^2=13 x^2+(mx+4m+7)^2=13 x^2+m^2x^2+16m^2+49=13

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x^2(1+m^2)+16m^2=36

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2x^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
 one year ago
Best ResponseYou've already chosen the best response.0x^2 + m^2(x+4)^2 +49 +14m(x+4) = 13 where did u get the 14m(x+4) part frm?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2gotta 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
 one year ago
Best ResponseYou've already chosen the best response.2and (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
 one year ago
Best ResponseYou've already chosen the best response.0which wolf are u talking about...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2mx + 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
 one year ago
Best ResponseYou've already chosen the best response.0shouldnt it be mmxx+ 8mm+14 im confuse ...were u adding it?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2that is NOT how you multiply

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i should prolly ask how you ahve been solving these before i came in to muddy the waters

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444263458406:dw what is the value of n?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2n^2 = 4^2+7^213 = 52 n = sqrt(52) right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2now we know the radius of a circle centered at 4,7 and can equate 2 circles to find where they meet (x+4)^2 + (y7)^2  52 = x^2 + y^2 13

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i will be back in the next 20 second ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i got it figured out :) ill be back as well ... food is on the table

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 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...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444264864450:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0o ok looks interesting

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2so, 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
 one year ago
Best ResponseYou've already chosen the best response.2n = sqrt(52) so we our circle of radius sqrt(52) centered at 4,7 is (x+4)^2 + (y7)^2 = 52 and equate this with the given: x^2 + y^2 = 13 (x+4)^2 + (y7)^2 +13 = 52 +x^2 + y^2 x^2+8x+16 + y^214y+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
 one year ago
Best ResponseYou've already chosen the best response.2this 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444265497662:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not sure if i follow all this ..but x^2+(4x+13)^2/49=13 49x^2+16x^2+104x+169=637 65x^2+104x468=0 5x^2+8x13=0 5x^25x+13x13=0 5x(x1)+13(x1)=0 5x+13=0 x=13/5 and x1=0 x=1

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2d^2 = (a^2+b^2), for some point (a,b) n^2 = a^2+b^2  k, for some given r^2=k (xa)^2 + (yb)^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
 one year ago
Best ResponseYou've already chosen the best response.2x=18/5 and x=2 i think you might have made a few errors

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2so it does generalize into a line solution of y = (ax+k)/b y = (4x+13)/7 in this case

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2x^2 + (4x+13)^2/49 = 13 49x^2 + 16x^2 +169 +8(13)x = 13(49) 65x^2 +104x 468 = 0 13 (x2) (5x+18) = 0

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2using 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
 one year ago
Best ResponseYou've already chosen the best response.2or as noted, y' = x1/y1 will give us the slopes as well

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg 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
 one year ago
Best ResponseYou've already chosen the best response.2you have to determine your points of tangency ... we found the x values (18/5 and 2), what are the y values ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y=m1(x+4)+7 when x=18/5 y=m1(18/5+4)+7 y=m1(2/5)+7 am i going right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2no, 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
 one year ago
Best ResponseYou've already chosen the best response.2the slope of that tangent at 2,3 is: 2/3 , the normal slope is 3/2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2y = m(xa)+b x^2 + m^2 (x^2 2ax + a^2) + b^2 +2mb(xa) = 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(bam) x + (am)^2 + b^2  2mba k = 0 this has one root value if the discriminant is equal to 0 soo (2m(bam))^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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2(m(xa)+b)^2 x^2 + m^2 (x^2 2ax + a^2) + b^2 +2mb(xa) = 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(bam) x + (am)^2 + b^2  2mba k = 0 (2m(bam))^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 (bam)^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^2k) m^2 4(2ab) m +4(ab^2k) divide off the 4 (a^2) m^4 (2ab) m^3 +(a^2+ab^2k) m^2 (2ab) m +(ab^2k)  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
 one year ago
Best ResponseYou've already chosen the best response.2might be some calc errors in there cause the wolf says we obtain a quadratic with your a,b

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2oh, thats just the 4ac part ... negate and add in the b^2 might help

amistre64
 one year ago
Best ResponseYou've already chosen the best response.24m^2 (bam)^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^2k) m^2 +4(2ab) m 4(ab^2k) (a^2+ab^2b^2k) m^2 +(2ab) m (ab^2k) \[m=\frac{ab\pm\sqrt{a^2b^2(a^2+ab^2b^2k)(ab^2k)}}{a^2+ab^2b^2k}\] not too sure how useful this would be ... doesnt look simple to memorize :) \[m=\frac{7(4)\pm\sqrt{16(49)(164(49)4913)(4(49)13)}}{164(49)4913}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2id stick with that: y = (ax+k)/b and plugging that into the circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok y do we have to do this part ...what are we finding for ? frm my last post to this post...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.2if you know of some other way ... feel free

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no urs is ok... ill figure it out... so whats the next step... and i really appreciate ur help

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2thru 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444270137860:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok like this dw:1444266705960:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2yes, 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
 one year ago
Best ResponseYou've already chosen the best response.2works coming early in the morning so i have to wrap this up soon

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea good up to there

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2since we know our point of tangency (x,y) our slope of tangency is just x/y (and slope of normalcy is y/x)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2we then construct our lines using our tangency points and the slopes required. yb = m(xa) for tangent points (a,b) and slope m

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2for the point (2,3) , tangent slope is 2/3 and normal slope is 3/2 lines are: tan y = 2/3 (x2)+3 norm y = 3/2 (x2) +3

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2it 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
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.2yep, 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
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.0ok thanks ..i will go through back all the post to make sure i understand everything

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2my 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
 one year ago
Best ResponseYou've already chosen the best response.2use the points of tangency to define slopes and lines as needed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think my problem here is ..i dont fully understand this tangency thing ...i will watch some videos on it..
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