Finding intersection of line and circle.

1. The problem statement, all variables and given/known data

I have a line obtained from using a slope of 1 and point (-sqrt(2),sqrt(2)):

y - sqrt(2) = 1 (x+sqrt(2))
y = 2*sqrt(2) + x

and a circle with radius 5 centered at origin.


My thought was to solve this parametrically



The line is the tangent line (blue line). I am trying to find the insersections (points) where the blue line meets the red circle.


2. Relevant equations



3. The attempt at a solution

Given point (-sqrt(2),sqrt(2)) and knowing the vector <1,1>

X = -sqrt(2)+t
Y = sqrt(2)+t

X = 5 * cos(t)
Y = 5 * sin(t)

Subsituting:

-sqrt(2)+t = 5 * cos(t)
sqrt(2)+t = 5 * sin(t)

I then attempt to solve for t, but cannot find a solution.

 

Adding and squaring what?

 

How it relates to any mathematical formula or understanding, taking your reply literally :


(-sqrt(2)+t)^2 + (sqrt(2)+t)^2 = (5 * cos(t))^2 + (5 * sin(t))^2

Yielded nothing useful.

 

http://www.wolframalpha.com/input/?...qrt(2)+t)^2+=+(5+*+cos(t))^2+++(5+*+sin(t))^2

Gives me t= ± sqrt(21/2)

Pluggin this into

X= 5*cos(t), gives ~ -4.97

Of which according to maple the approx points of X I am looking for are 1.826 and -4.654

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If you would please explain to me the mathematical formula you were apply when squaring both and adding both equations

Edit:

Using

X = -sqrt(2) ± sqrt(21/2) does indeed yield the points of X I am looking for, however I am a bit confused why the parametic equation from the circle does not.

I am not looking for a cut and dry tell me the solution answer. I am more interested in actually -learning- how it was done. At this point I don't understand why you did what you did worked.

 

[itex]
\left( -\sqrt {2}+t \right) ^{2}+ \left( \sqrt {2}+t \right) ^{2}=25
\, \left( \cos \left( t \right) \right) ^{2}+25\, \left( \sin \left(
t \right) \right) ^{2}
[/itex]

[itex]
\left( -\sqrt {2}+t \right) ^{2}+ \left( \sqrt {2}+t \right) ^{2}=25(
\, \left( \cos \left( t \right) \right) ^{2}+\left( \sin \left(
t \right) \right) ^{2})
[/itex]


[itex]
\left( -\sqrt {2}+t \right) ^{2}+ \left( \sqrt {2}+t \right) ^{2}=25(1)
[/itex]

[itex]
\left( -\sqrt {2}+t \right) ^{2}+ \left( \sqrt {2}+t \right) ^{2}=25
[/itex]

[itex]
(2+2\sqrt{2}t+t^{2})+(2-2\sqrt{2}t+t^{2}) = 25
[/itex]

[itex]
4+2t^{2} = 25
[/itex]

[itex]
{t}^{2}=\frac{21}{2}
[/itex]

[itex]
{t}=±\sqrt{\frac{21}{2}}
[/itex]

I understand basically in the most simplistic sense you did

x(t)^2+y(t)^2 = x1(t)^2+y1(t)^2

Which will should still give a t value that satisfies what I needed, you squared it for the trig identity cos^2 + sin^2 = 1 to cancel out the trig, which messed up my equation when trying to solve for it. It seems there was no known mathematical procedure or formula that unique or specific to this type of problem but rather intuition was used to make the problem solvable, which was not as obvious to me.

Still for this value of T you find using x(t)^2+y(t)^2 = x1(t)^2+y1(t)^2 it doesnt seem to provide

x(t) = x1(t) and y(t)=y1(t)

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Am I correct in my understanding?