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July 26th, 2010, 02:03 AM   #1
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Tangents for a Circle

I have found this exercise on the internet, but I just can't figure it out. I tried it with vectors

Quote:
 Find the equations of the two lines which pass through the point (0,4) and form tangents to a circle of radius 2, centred on the origin.
I know the equation of the circle: x^2 + y^2 = 4

But how should I continue?

 July 26th, 2010, 02:55 AM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Tangents for a Circle My first instinct would be to draw the circle, the point, and the tangent lines. Use Pythagorean theorem to get the length of the segments (ie sqrt20). You should be able to see that this problem is identical to finding the intersection of two circles: A- centered at the origin, radius 2, standard equation x^2 + y^2 = 4 B- centered at (0,4), radius sqrt20, standard equation ______ Solve this system of equations and you're done!
 July 26th, 2010, 03:50 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,924 Thanks: 2203 No, you're not, as the question asked for the equations of the tangents, not their points of contact. Also, your calculated radius was incorrect. Each equation must have the form y = mx + 4. Substituting for y in the equation x² + y² = 4 gives x² + (mx + 4)² = 4, i.e., (m² + 1)x² + 8mx + 12 = 0. This equation's discriminant must be 0, so 16m² - 12(m² + 1) = 0, i.e., m² = 3. Hence the tangents' equations are y = ?3x + 4 and y = -?3x + 4.
July 26th, 2010, 05:23 AM   #4
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Re: Tangents for a Circle

Hello, much12!

Did you make a sketch?

Quote:
 Find the equations of the two lines which pass through the point (0,4) and form tangents to a circle of radius 2, centered on the origin.
Code:
              A o (0,4)
/|\
/ | \
/  |  \
/   |   \
/    |    \
/   * * *   \
/*     |     *\
C o       |       o B
*  *     |     *  *
*   |   * 2
*       * | *       *
- - * - - - - *O- - - - * - -

$\text{We have: }\:\angle ABO= 90^o,\;AO = 4,\;BO = 2$

[color=beige]. . [/color]$\text{Hence: }\:\angle A\,=\,30^o$

$\text{Then }AB\text{ has slope }-\sqrt{3},\,\text{ and }AC\text{ has slope }+\sqrt{3}.$
[color=beige]. . [/color]$\text{and both have a }y\text{-intercept of 4.}$

$\text{The equations are: }\;\begin{Bmatrix} y=&-\sqrt{3}x\,+\,4 \\ \\ \\ \\ y=&\sqrt{3}x\,+\,4 \end{Bmatrix}=$

 July 26th, 2010, 06:12 AM #5 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Tangents for a Circle Guess I should have drawn the picture... SQRT(12)! And while the form "must" be y = mx + 4, the general case requires using the distance between the centers to find the radius of the second, and then solving a system of equations. (algebraically)
 July 26th, 2010, 08:10 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,924 Thanks: 2203 There's nothing particularly special about this case, where it wasn't necessary to find the distance you refer to.
 July 26th, 2010, 08:48 AM #7 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Tangents for a Circle I thought the centers both being on the y-axis was kinda special!
 July 26th, 2010, 01:38 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,924 Thanks: 2203 That makes your initial distance calculation simpler, but has no significant effect on the method I used.

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