Quandaries and Queries
Could you please help me with the following problem:
I completed the squares to put the equation in standard form and got
(x-3)2 + (y-3)2 = 1
Hence this is the circle with centre (3,3) and radius 1. The line y=mx is a line through the origin. This information allowed me to draw a diagram and see the two possible tangent lines. On the diagram P is one of the points of tangency, I called it (a,b), C is the centre of the circle (3,3) and O is the origin.
Since PO is tangent to the circle, angle CPO is a right angle and thus Pythagoras theorem ensures that
|CP|2 + |PO|2 = |OC|2
1 + a2 + b2 = 18
Since P is on the circle,
a2 + b2 - 6a -6b + 17 = 0
Substituting from the equation above gives
6a + 6b = 34
b = 17/3 - a
If you now substitute this value for b into a2 + b2 - 6a -6b + 17 = 0 you will have a quadratic in a which you can solve for a.
Finally find b from b = 17/3 - a and the unknown slope is b/a