by Carlhole » Sun 28 Oct 2007, 17:33:25
$this->bbcode_second_pass_quote('jasonraymondson', 'P')MS, your the math teacher, can we get some help with this problem?
You still need help? Dude! Put the math book down and slowly back away!
You've heard of the Pythagorean Theorem, right?
a^2 + b^2 = c^2
...look at the right triangle in the above diagram. (Instead of sin t, substitute Y. Instead of cos t, substitute X).
The hypotenuse of the triangle is always equal to 1.
So according to the Pythagorean Theorem, x^2 + y^2 = 1 (your original equation).
Now imagine the point where the triangle interesects the circle travels around a full revolution. The x and y legs of the right triangle vary but, of course, the hypotenuse always remains constant at 1.
So the equation, x^2 + y^2 = 1, defines a unit circle (circle with hypotenuse of length 1) whose center lies on the origin.
The link in my previous post shows the general equation for circles whose centers are located at point (h,k) and whose radius is r. And there is a graph there where you can plug in different values. In your case, h and k are both zero and r = 1.
The circle defined by your equation passes through point (1,0) on the X-axis. That is the closest point to (2,0).
Problem solved.