Question

Find the equation of the tangent line at (-1,5) on the circle x^2+y^2+2x-9=0

Answer 1

Do you know how to do implicit differentiation?

Answer 2

take the derivative of each term with respect to x

use these rules:
\[ \frac{d}{dx} x = \frac{dx}{dx} =1 \\ \frac{d}{dx} y= \frac{dy}{dx} \]

Answer 3

no :(

Answer 4

can you do
\[ \frac{d}{dx} x^2 \]?

Answer 5

substitute it to (-1,5)?

Answer 6

no, using the (-1,5) comes at the end.

First we have to find the slope of the tangent line to the curve.
that means we find dy/dx (which by definition is the slope of the tangent line)
and *that* means use differential calculus

you need to first take the derivative of your function.

The first step is, can you find the derivative of x^2 with respect to x?

Answer 7

wait

Answer 8

\[******x ^{2}+y^{2}+4x-9=0 ******\]

Answer 9

@phi (x+2)^2?

Answer 10

it looks like you completed the square, which is a way to put the equation into the standard form of a circle. That is useful for finding the center of the circle.

But, it is not needed for this problem.

Answer 11

this might actually not be a calculus problem
i have seen it done without using derivatives
it is a pain, derivatives make it much easier
but if this is not calc class then this will make no sense

Answer 12

oh! @SeniorHigh what course is this ??

Answer 13

introduction to analytical geometry&calculus

Answer 14

Have you learned (or were you supposed to have learned) how to take a derivative?

Answer 15

we didn't discuss it yet.. still on the circles (CONICS)

Answer 16

thought so
damn

Answer 17

In that case, we should plot the circle and the point (-1,5) because we are going (painful) geometry

Answer 18

i think there is some method arguing that when you solve the simultaneous equation the discriminant must be zero if they are to touch
let me see if i can remember it

Answer 19

sir what is the center and the slope of radius?

Answer 20

oh no it is not that bad at all

Answer 21

I can see how to do it using ratios of similar triangles.

1) the radius will form a right angle with the tangent line

Answer 22

1) find the center of the circle by completing the square
2) find the slope of the line from the center to the point \((-1,5)\)
3) the line will be perpendicular, take the negative recprocal
4) use the point slope formula

Answer 23

@phi seem reasonable?

Answer 24

we can walk through it if you like, i am fairly certain that this will work

Answer 25

(-1,5) is not on the circle

Answer 26

oh damn

Answer 27

i must have read the problem wrong, thought it said "one the circle"

Answer 28

but we do have similar right triangles.
we know length 5, "height" (i.e. radius) sqr(10), and hypotenuse sqr(15)

Answer 29

do i need to perform the completing the square in \[x ^{2+y ^{2}+4x-9=0}\]?

Answer 30

Yes, to do this problem you need to at least know its center and radius

can you complete the square on the x terms?

Answer 31

\(4x\) or \(2x\) ?

Answer 32

4x @satellite73

Answer 33

that shoots all my calculations to hell

Answer 34

\[(x+2)^{2}+(y-3)^{2}\] is this right?

Answer 35

you sure? in the question you wrote
\[x^2+y^2+2x-9=0\]

Answer 36

Gack! the 4x makes it harder!

and, there is now y to the first power, so you do not complete the square on the y part of the problem

Answer 37

ikr

Answer 38

typo @satellite73

Answer 39

i know! that's why i hate math T.T

Answer 40

so far i am at
\[y^2=13-(x+2)^2\\
y=m(x+1)-5\]

Answer 41

now i will try to square the bottom one
try to solve for \(x\) but only set the discriminate equal to zero because there should only be one solution, then solve that equation for \(m\)

Answer 42

there will be two tangent lines

Answer 43

yes, when i solve for \(m\) it will be a quadratic, should get two solutions

Answer 44

meanwhile, did @SeniorHigh find the equation of the circle ?

Answer 45

this would be a ton easier if it was \(x^2+y^2+2x-9=0\)

Answer 46

i don't know how to

Answer 47

but according to my classmate x+5-24=0

Answer 48

Remember we are trying to form an equation of the form
\[ (x-a)^2 + (y^2 - b^2) = r^2 \]
Let's complete the square on the x part
\[ x^2+4x+y^2−9=0 \\ x^2+4x+y^2=9 \]
first, notice there is no y term. we don't have to do anything for the y^2 term
next, notice the 4x term. we divide the 4 by 2. 4/2= 2. square it. 2*2= 4
that means we want a +4. add +4 to both sides
\[ x^2+4x+4 +y^2=9+4 \]
the x^2+4x+4 = (x+2)^2 (multiply out to show it's true)
\[ (x+2)^2 +y^2= 13\]

Answer 49

based on that equation, can you identify the center of the circle and its radius?

Answer 50

yes but can you explain how to do it?

Answer 51

yes what?

Answer 52

I think you mean no, can you explain it?

match the equations
\[ (x+2)^2 +y^2= 13 \\ (x-a)^2 + (y^2 - b^2) = r^2 \]
the center will be (a,b)

we have x+2 but that does not match x-a. but we can write it as
(x- -2)
we have just y^2, but we could write it (y-0)^2. we don't have an r^2, but we can write
\( 13= (\sqrt{13})^2 \)

Answer 53

so match these equations
\[ (x - -2)^2 +(y-0)^2 = (\sqrt{13})^2 \\ (x-a)^2 + (y-b)^2 = r^2 \]
what is a and b? (the center will be (a,b) )
what is r ?

Answer 54

center is -2,0 and r is 5

Answer 55

yes, -2,0 for the center. r is not 5. you match up r^2 with (sqr(13))^2
in other words, r matches sqr(13)

Answer 56

oohh.. I get it now.. thank you :)

Answer 57

you should plot (or sketch) the problem.
Here it is

Answer 58

we can find the lengths of the legs of the right triangle.

1) the length of the "short leg" is the radius. we know that is sqr(13)
2) what is the distance from the center to point A ? this is the hypotenuse
3) from 1) and 2) we can find the 3rd leg using Pythagoras

Answer 59

you should find the "sides" are sqr(13) and the hypotenuse is sqr(26)
in other words a 45-45-90 triangle
I can see how to use trig (tangent of sum of two angles) to find the slope of the tangent line.
There are probably other approaches

Answer 60

Here is a picture

Answer 61

the last step to find the equation of the line is
y-5 = m(x - -1) or
y-5= m(x+1)
where m is the slope, or the perpendicular -1/m (for the other tangent line)