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Coordinate Geometry of Circles

Date: 02/26/2003 at 16:04:05
From: Anas
Subject: Coordinate Geometry of Circles

The line with equation y = mx is a tangent to the circle with equation 
x^2 + y^2 - 6x -6y +17 = 0. Find the possible values of m.

Date: 02/26/2003 at 16:12:19
From: Doctor Schwa
Subject: Re: Coordinate Geometry of Circles

Hi Anas,

I drew a picture of the circle (I got its center as (3,3), and radius 
1, is that right?) on the coordinate plane.

Then I'm looking for lines through the origin that are tangent to the
circle.  I can see from my drawing that there should be two of them.

Where the line touches the circle is perpendicular to the radius 
(that's what it means to be tangent to a circle). So I get a right 
triangle. The hypotenuse is the distance from the origin to the center 
of the circle; one leg is the radius; so I can solve for the other 

That other leg gives the radius of another circle, centered at the 
origin; then I need to find the intersection points of the two circles 
in order to find the slope.

Alternatively, rather than using the length of that side, I could use 
inverse sin to find the angle measure, and then use that to find the 

Do those hints help?

- Doctor Schwa, The Math Forum 

Date: 02/27/2003 at 13:51:38
From: Doctor Wilkinson
Subject: Re: Coordinate Geometry of Circles

If you substitute mx for y in the equation of the circle, you'll get a 
quadratic equation in x:

  (1+m^2)x^2 - (1+m)6x + 17 = 0

You can use the quadratic formula to get 

      -[-(1+m)6] +/- sqrt([-(1+m)6]^2 - 4[1+m^2][17])
  x = -----------------------------------------------

Now, what does this tell you?  If you choose a value for m, it tells
you the x-coordinate of the points where y = mx intersects the circle. 

For most values of m, the number of intersections will be either zero
(i.e., the line misses the circle entirely) or two (i.e., the line
cuts through the interior of the circle.  But if the line is tangent
to the circle, then there is only one intersection.  And for there to
be only one intersection, the term

  sqrt([-(1+m)6]^2 - 4[1+m^2][17])
must be equal to zero.  Find the value(s) of m for which this is true,
and you'll know the slopes of the lines you're looking for. 

- Doctor Wilkinson and Doctor Ian, The Math Forum 
Associated Topics:
College Conic Sections/Circles
High School Conic Sections/Circles

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