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

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

, which on equating discriminant to be zero, gives

$[3 (m + 1 {)]}^{2} - 17 (m^{2} + 1) = 0$

and just solve this for $m$ .

Method 2: The distance of $O$ from the centre of the circle is $3 \sqrt{3}$ , so the tangent has length $\sqrt{17}$ . Hence the gradient is $\tan (π / 4) \pm \tan^{- 1} (1 / \sqrt{17})$ (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.