Question

Use the discriminant to find the gradients of the lines that
pass through the point (7, 1) and are tangent to the circle
x^2 + y^2 = 25.

Answer 1

@satellite73 help please ^^

Answer 2

ok, we need \(y'\) using implicit diff you get
\[2x+2yy'=0\] solve for \(y'\) gives \(y'=-\frac{x}{y}\) and now plug in the numbers

Answer 3

you could also solve for \(y\) in this example, and get
\[y=\pm\sqrt{25-x^2}\] then take the derivative, then plug in the numbers
try it you will see you get the same answer

Answer 4

y' = dy/dx right?

Answer 5

and the third method is to recognize this as a circle and know that the answer is always \(-\frac{x}{y}\) for a circle

Answer 6

yeah, \(y'\) takes two keystrokeds, \(\frac{dy}{dx}\) takes 12

Answer 7

ok the question asks to use the discriminant.. which way would that be?

Answer 8

i have absolutely no idea
the only definition i know for "discriminant" is \(ax^2+bx+c=0\) then the discriminant is \(b^2-4ac\)

Answer 9

you are being asked for the slope of the line tangent to the circle
idea is to find the derivative, plug in the numbers

Answer 10

(7, 1) and a tangent to the circle x^2 + y^2 = 25
Is it possible to find the equation of this line?

Answer 11

sure

Answer 12

you have a point, so all you need is the slope

Answer 13

is it clear how to find the slope, or no?

Answer 14

yep all clear

Answer 15

slope is 25/28?

Answer 16

ooh lets back up

Answer 17

oh crap i read the problem wrong
my fault
\((7,1)\) is not a point on the circle is it? you want the equation for the line tangent to the circle through the point \((7,1)\)

Answer 18

yea i think so

Answer 19

a point on the circle will look like
\[(x, \sqrt{25-x^2})\] and the slope will be
\[\frac{\sqrt{25-x^2}-1}{x-7}\]

Answer 20

this in turn must be equal to the derivative, which is
\[-\frac{x}{\sqrt{25-x^2}}\]

Answer 21

set them equal and solve for \(x)

Answer 22

hmm?

Answer 23

you have a point on the circle
it looks like \((x,\sqrt{25-x^2})\) or \((x,-\sqrt{25-x^2})\)

Answer 24

then you have a point not on the circle, \((7,1)\)

Answer 25

ok i get that

Answer 26

the slope between those two points is going to be
\[\frac{\sqrt{25-x^2}-1}{x-7}\]

Answer 27

ok

Answer 28

maybe it would be clearer if i replaced \(x\) by \(a\) but lets keep with the \(x\)

Answer 29

now we take the derivative of
\[y=\sqrt{25-x^2}\]

Answer 30

and get
\[y'=\frac{-x}{\sqrt{25-x^2}}\]

Answer 31

ok so far?

Answer 32

yep

Answer 33

now you want the line through \((7,1)\) tangent to the circle at some point
if it is tangent to the circle, then the slope must be equal to the derivative at the point where it touches the circle

Answer 34

im pretty sure there would be a quicker way because they rest of the question are pretty basic :/ and not this complex lol

Answer 35

in other words \(\frac{\sqrt{25-x^2}-1}{x-7}\) must be equal to \(\frac{-x}{\sqrt{25-x^2}}\)

Answer 36

yeah i am thinking the same thing, but i don't see it

Answer 37

maybe solve two equation
\[x^2+y^2=25\]
\[\frac{y-1}{x-7}=-\frac{x}{y}\]

Answer 38

yeah, that might be easier !!

Answer 39

here is the worked example. i think its similiar

Answer 40

does the solution look similar to the two equations i wrote above?

Answer 41

yeah same idea

Answer 42

by the way these equations are actually identical to my complicated method, because \(y=\sqrt{25-x^2}\) but i can see that this will be much easier

Answer 43

ok i got a answer! ^^ Thank you so much for your time and your effort!!

Answer 44

lol what is it?

Answer 45

i thought i got it but i ended up with a negative square root :(

Answer 46

so i did y-1 = m(x-7) first right?

Answer 47

ooh i see
learn something new every day! in order for the line to be tangent, it means it must touch
therefore the discriminant must be zero!! now i see what you were saying at the beginning!!!

Answer 48

then i got y = mx-7m+1

Answer 49

yeah i am there too

Answer 50

now do i have to square that? cause that will be messy

Answer 51

now i am at the step where you set
\[(mx-7m+1)^2+x^2=25\]

Answer 52

well im going to expand it and see what i get :/ this is annoying

Answer 53

yeah i am going to do it as well

Answer 54

\[m^2 x^2-14 m^2 x+49 m^2+2 m x-14 m+1+x^2=25\]

Answer 55

hehe, im gonna double check my answers with wolfram ^^

Answer 56

double check? i just used wolfram!

Answer 57

now lets get the quadratic equation in \(x\)

Answer 58

lol... i did it on paper first :P

Answer 59

\[(m^2+1)x^2+(-14m^2+2m)x+(49m^2-14m-24)=0\] really??

Answer 60

i thought this was supposed to be the easy way!!

Answer 61

now we set \(b^2-4ac=0\) right?

Answer 62

yea... wow... this is no easy way

Answer 63

\((-14m^2+2m)^2-4(m^2+1)(49m^2-14m-24)=0\)

Answer 64

i think i messed up somewhere
i bet it is actually easier to use the derivative

Answer 65

Maybe their is a easier way using the discriminant...

Answer 66

anyways ill ask my teacher tomorrow and see what she says...
Thanks for your help!!!

Answer 67

sorry i will think about it and get back to you

Answer 68

If if you dont have time, dont worry about it :)

Answer 69

I got to sleep its 1:25 in the morning

Answer 70

ok if you see this when you get up, it is not that hard at all
solve
\[x^2+y^2=25\\
\frac{y-1}{x-7}=-\frac{x}{y}\] and get
\[x=3,y=4\] or
\[x=4,y=-3\]