finding the equation of the tangent line to a circle

• Feb 18th 2010, 04:41 PM
alessandromangione
finding 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 PM
skeeter
Quote:

Originally Posted by alessandromangione
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 $r^2(1+m^2)=b^2$, but I can tell you that the slope of the tangent line is perpendicular to the slope of the radius to the point $(1 , 2\sqrt{2})$

slope of the radius is $\frac{2\sqrt{2}}{1}$

tangent slope would be $-\frac{1}{2\sqrt{2}}$

tangent line equation is ...

$y - 2\sqrt{2} = -\frac{1}{2\sqrt{2}}(x-1)$
• Feb 18th 2010, 08:50 PM
alessandromangione
Quote:

Originally Posted by skeeter
I'm not familiar with the equation $r^2(1+m^2)=b^2$, but I can tell you that the slope of the tangent line is perpendicular to the slope of the radius to the point $(1 , 2\sqrt{2})$

slope of the radius is $\frac{2\sqrt{2}}{1}$

tangent slope would be $-\frac{1}{2\sqrt{2}}$

tangent line equation is ...

$y - 2\sqrt{2} = -\frac{1}{2\sqrt{2}}(x-1)$

well the result should be square root of 2 +4y-9square root of 2=0....maybe someone else can help me out?
anyway thank u !
• Apr 12th 2010, 09:26 PM
lsp2010
I 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 PM
sa-ri-ga-ma
Quote:

Originally Posted by alessandromangione
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

In the given problem r = 3.
Differentiating the equation of the circle, you get m. Thus
2x + 2y*y'= 0 Put the values of x and y and find y' = m.
Using the formula
r^2(1+m^2) = b^2, find b.
Then the equation of the tangent is
y = mx + b.
• Apr 13th 2010, 10:20 AM
earboth
Quote:

Originally Posted by alessandromangione
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

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 $T\left(\frac{-bm}{1+m^2} , \frac{b}{1+m^2}\right)$

3. Consequently you'll get a system of equations:

$\left|\begin{array}{rcl}\dfrac{-bm}{1+m^2}&=&1 \\ \\ \dfrac{b}{1+m^2}&=&2\sqrt{2} \end{array}\right.$

4. Solve this system for b and m and plug in these values into the equation of a straight line.
Spoiler:
I've got $y = -\frac14 \sqrt{2} \cdot x + \frac94 \sqrt{2}$