seeing that r^2(1+m^2)=b^2...use this method to find an equation of the tangent line to the circle x^2+y^2=9 at the point ( 1, 2square root of 2 )
i'm totally lost...thanks in advance
I'm not familiar with the equation $\displaystyle r^2(1+m^2)=b^2$, but I can tell you that the slope of the tangent line is perpendicular to the slope of the radius to the point $\displaystyle (1 , 2\sqrt{2})$
slope of the radius is $\displaystyle \frac{2\sqrt{2}}{1}$
tangent slope would be $\displaystyle -\frac{1}{2\sqrt{2}}$
tangent line equation is ...
$\displaystyle y - 2\sqrt{2} = -\frac{1}{2\sqrt{2}}(x-1)$
1. If your equation is satisfied by r, m and b the line in question is a tangent (in Germany this equation is called "condition for the existance of a tangent to a circle")
2. The tangent point is $\displaystyle T\left(\frac{-bm}{1+m^2} , \frac{b}{1+m^2}\right)$
3. Consequently you'll get a system of equations:
$\displaystyle \left|\begin{array}{rcl}\dfrac{-bm}{1+m^2}&=&1 \\ \\ \dfrac{b}{1+m^2}&=&2\sqrt{2} \end{array}\right.$
4. Solve this system for b and m and plug in these values into the equation of a straight line.
Spoiler: