Stalker23 said:
no, sorry, we've never used the constant r- what is it. is it the radius of the circle- correlation coeficiant??
Now you've confused me! I've looked back through all the responses and find no reference to a "constant r"! Did someone edit or delete it?
Any way, the point is that the circle given, x^2- 4x+ y^2+ 3= 0, which is equivalent to (x- 2)^2+ y^2= 1 is a circle with center at (2, 0) and radius 1. It doesn't pass through (0,0) so you want to find two lines that do go through (0,0) and are tangent to the circle at two
other points.
You could do this geometrically: The line segement from (0,0) to (2, 0), the line segment from (2, 0) to the point of tangency, and the line segment from (0,0) to the point of tangency form a right triangle with one leg of length 1 (the radius of the circle) and hypotenuse of length 2 (from (0,0) to (2, 0)) and so the length of the other leg is \sqrt{3}. Any point (x,y) on the circle must satisfy (x-2)^2+ y^2= 1 and if the line from (0,0) to (x,y) has length \sqrt{3} then
also x^2+ y^2= 3. Solve those two equations for the points of tangency.
But, using calculus, do what I suggested before. The line from (0,0) to (x, y) has slope \frac{y}{x} and that must be the y' at the point on the circle. Using implicit differentiation, 2x- 4+ 2yy'= 0 so y'= \frac{4-2x}{2y}= \frac{2-x}{y}. Solve \frac{2- x}{y}= \frac{y}{x} along with x^2- 4x+ y^2+ 3= 0 for x and y. Those two methods should give the same point.