1. ## TANGENT and CIRCLE

I'm having a problem with the following question. If anyone can help, that would be very much appreciated.

(1) Suppose that (a,b) is on the circle (x^2)+(y^2)=(r^2). Show that the line ax+by=(r^2) is tangent to the circle at (a,b).

PLEASE, IF ANYONE CAN HELP, I WOULD LOVE IT.

2. after an implicit differentiation of the circles equation we get:

$\displaystyle \begin{gathered} 2x + 2y\frac{{dy}} {{dx}} = 0 \hfill \\ \Leftrightarrow \frac{{dy}} {{dx}} = - \frac{x} {y} \hfill \\ \end{gathered}$

thus the general equation of the slope of any tangent to the circle is:

$\displaystyle \frac{{dy}} {{dx}} = - \frac{x} {y}$

the general equation of a line passing through a coordinate (a,b) with a slope m is:

$\displaystyle y - b = m(x - a)$

the slope m can be obtained from:

$\displaystyle m = - \left. {\frac{x} {y}} \right|_{\left( {x,y} \right) = (a,b)} = - \frac{a} {b}$

thus the equation of the tangent is:

$\displaystyle \begin{gathered} y - b = - \frac{a} {b}(x - a) \hfill \\ \hfill \\ \Leftrightarrow by - b^2 = - ax + a^2 \hfill \\ \Leftrightarrow ax + by = a^2 + b^2 \hfill \\ \end{gathered}$

if we substitute (x,y) = (a,b) in the equation of the given circle we get:

$\displaystyle a^2 + b^2 = r^2$

thus the equation of the tangent is:

$\displaystyle ax + by = r^2$