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

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- Feb 18th 2010, 04:41 PMalessandromangionefinding the equation of the tangent line to a circle
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 - Feb 18th 2010, 05:36 PMskeeter

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 ...

$\displaystyle y - 2\sqrt{2} = -\frac{1}{2\sqrt{2}}(x-1)$ - Feb 18th 2010, 08:50 PMalessandromangione
- Apr 12th 2010, 09:26 PMlsp2010I need help! ):
I am completely stuck on this problem. please help me work it out. ):

Find a formula that gives the slope of a line tangent to the circle (x^2)+(y^2)=(r^2) at any point (in the first quadrant) (x,sqrt((r^2)-y^2)). - Apr 12th 2010, 10:13 PMsa-ri-ga-ma
- Apr 13th 2010, 10:20 AMearboth
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.

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