1. TANGENCY Example 17Prove that the line 2x + y = 19 is a tangent to the circle x 2 + y 2 - 6x + 4y - 32 = 0, and also find the point of contact. ******** 2x + y = 19 so y = 19 – 2x Replace y by (19 – 2x) in the circle equation. x 2 + y 2 - 6x + 4y - 32 = 0 x 2 + (19 – 2x) 2 - 6x + 4(19 – 2x) - 32 = 0 x 2 + 361 – 76x + 4x 2 - 6x + 76 – 8x - 32 = 0 5x 2 – 90x + 405 = 0 (  5) x 2 – 18x + 81 = 0 (x – 9)(x – 9) = 0 x = 9 Using y = 19 – 2x If x = 9 then y = 1 (9,1) is only point of contact so line must be a tangent ! NAB

2. Using Discriminants At the line x 2 – 18x + 81 = 0 we can also show there is only one solution by showing that the discriminant is zero. For x 2 – 18x + 81 = 0, a =1, b = -18 & c = 81 So b 2 – 4ac =(-18) 2 – 4 X 1 X 81= 364 - 364= 0 Since disc = 0 then equation has only one root so there is only one point of contact so line is a tangent. The next example uses discriminants in a slightly different way.

3. Example18 Find the equations of the tangents to the circle x 2 + y 2 – 4y – 6 = 0 from the point (0,-8). ******* x 2 + y 2 – 4y – 6 = 0 2g = 0 so g = 0. 2f = -4 so f = -2. Centre is (0,2) 2 -8 Y ie Each tangent takes the form y = mx -8 Replace y by (mx – 8) in the circle equation to find where they meet.This gives us … x 2 + y 2 – 4y – 6 = 0 x 2 + (mx – 8) 2 – 4(mx – 8) – 6 = 0 x 2 + m 2 x 2 – 16mx + 64 –4mx + 32 – 6 = 0 (m 2 + 1)x 2 – 20mx + 90 = 0 In this quadratic a = (m 2 + 1)b = -20mc =90

4. For tangency we need disc = 0 b 2 – 4ac = 0 (-20m) 2 – 4 X (m 2 + 1) X 90 = 0 400m 2 – 360m 2 – 360 = 0 40m 2 – 360 = 0 40m 2 = 360 m 2 = 9 m = -3 or 3 So the two tangents are y = -3x – 8 and y = 3x - 8 and the gradients are reflected in the symmetry of the diagram.