Tangent to a Circle
By Robbie Hamilton on Monday, February 04, 2002 - 09:05 pm:

Given the equation of a circle x2 +y2 -6x-6y+17=0 how do you find the tangents to the circle with equation y=mx?

By Kerwin Hui on Monday, February 04, 2002 - 10:38 pm:

Hint: A line is tangent to a circle if and only if it intersects the circle at only one point. Use this fact to equate a discriminant to be zero.

Alternatively, you know the centre of the circle is at (3,3) and it's radius is 1. Use a bit of geometry.

Kerwin

By Yatir Halevi on Tuesday, February 05, 2002 - 01:32 pm:

Or Calculus, if you know...

Yatir

By Robbie Hamilton on Wednesday, February 06, 2002 - 04:45 pm:

yes I can calculate the gradient of the cirle,but I was looking for an elegant solution connecting the gradient of the circle with the gradient of the line. All my solutions are messy and longwinded - not suitable for teaching to A level students.

By Kerwin Hui on Wednesday, February 06, 2002 - 10:58 pm:
Method 1: Substitute y=m x into the equation yields (m2+1)x2-6(m+1)x+17=0, which on equating discriminant to be zero, gives

[3(m+1)]2-17(m2+1)=0

and just solve this for m.

Method 2: The distance of O from the centre of the circle is 3Ö3, so the tangent has length
  __
Ö17
 

. Hence the gradient is
tan(p/4)± tan-1(1/   __
Ö17
 
 )

(draw yourself a picture to see why). Evaluate this using the appropriate formula.

Kerwin

By Robbie Hamilton on Thursday, February 07, 2002 - 08:05 am:

Thank you very much, just what I needed.