tangent: If there is a cycle with a central point (1, 2) and

[kristina]

If you be so kind to help me, my problem is this:

If there is a cycle with a central point S(1, 2) and radius r = 2, can you give me a generale formula to find a tangent on that cycle?

I did this: if radius is 2 than : 2+1=3 and -2+1= -1. So t=3 and t=-1 (this is correct),but I` do not know is this the right way to get them ,maybe, I` got this by chance.

Because of that I need to know the general formula or procedure, how to get a tangent, if I know ,a r,and S. Please help me!!!

 

[pka]

Re: TANGENT

If there is a cycle(circle) with a central(center) point: S(1,2) and r(radius)=2. Please can you give me a general formula to find a tangent on that cycle(circle).

I will assume that is a questions about circles.
In this case the equation of this circle is \(\displaystyle \left( {x - 1} \right)^2 + \left( {y - 2} \right)^2 = 4.\)
That means that if (p,q) is a point on the circle then \(\displaystyle \left( {p - 1} \right)^2 + \left( {q - 2} \right)^2 = 4.\)
Thus the tangent to the circle at (p,q) is perpendicular to the line determined by the points (p,q) and (1,2). So the slope of the tangent is \(\displaystyle \frac{{ - \left( {p - 1} \right)}}{{\left( {q - 2} \right)}},\quad \left( {q \not= 2} \right).\)