The Student Room Group
rewrite it in the form (x+a)²+(y+b)²+c=0
√c is the radius
(-a,-b) is the centre

Draw out a little picture and it becomes clear that the two possible tangents are at (0,2) or (3,-1), as the line y=mx must pass through the origin.

You can then work out the gradients using
(y one - y two)/(x one - x two) - cant work out how to get subscript...

HTH
Reply 2
Fenchurch
A line L has the equation y=mx and a circle C has equation x²+y²-6x-4y+9=0

a) Given that L is a tangent to C find the possible values of m
b) Find the range of values of m given that L intersects C in two distinct points
c) Find the range of the valuse of m given that L and C do not intersect.


completing the square on x²+y²-6x-4y+9=0
is

(x-3)² -9 + (y-2)² -4+9=0
(x-3)²+(y-2)²=4 centre = (3,2)

when x=0

(-3)²+(y-2)²=4
9+y²-4y+4=4
y²-4y+9=0
(y-2)² =2±√5 so 0,2±√5 is one set of co-ordinates

(x-3)²+(y-2)²=4 when y=0

(x-3)²+(-2)²=4
(x-3)²=0
(3,0) for co-ordinate 2

not sure about the value m

Latest