Also, can anyone prove that the radius which shares the same point of tangency is perpendicular to the tangent line.
"A tangent is perpendicular to the radius that shares the point of tangency."
Thanks alot!
Take a paper and draw the following:
Let the centre of the circle be O. Let PQ be a tangent touching the circle at T. Clearly OT is a radial line (radius).
If OT was not perpendicular to PQ, then we can drop a perpendicular from O onto PQ and call the foot of the perpendicular S. Thus we see that OS (the perpendicular) is shorter than OT (some other line joining O and a point on PQ). But is that possible?
There is one theorem in circle.
If you draw a chord AB from the point of contact (A) of the tangent to the circle, then the angle subtended by the chord at any point P on the circumference (angle APB ) is equal to the angle between chord and the tangent.
Now suppose the chord AB is the diameter of the circle. Then angle APB = π/2. So the angle between AB ( i.e. radius ) and the tangent is π/2
Calclus makes this really easy.
A line from the center $\displaystyle (x_0,y_0)$ to a point $\displaystyle (x,y)$ on the circoe Has equation $\displaystyle y-y_0=\frac{y-y_0}{x-x_0}(x-x_0)$. Differentiating the equatio of a circle we get : $\displaystyle \frac{dy}{dx}= -\frac{x-x_0}{y-y_0}$.