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July 26th, 2010, 02:03 AM  #1  
Newbie Joined: Jul 2010 Posts: 1 Thanks: 0  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:
But how should I continue? Thanks in advance!!  
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  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Tangents for a Circle Hello, much12! Did you make a sketch? Quote:
Code: A o (0,4) /\ /  \ /  \ /  \ /  \ / * * * \ /*  *\ C o  o B * *  * * *  * 2 * *  * *   *     *O    *   [color=beige]. . [/color] [color=beige]. . [/color]  
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 yaxis 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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