Quandaries
and Queries 

Hello, Could you please help me with the following problem: Thank you, 

Hi Lech, I completed the squares to put the equation in standard form and got (x3)^{2} + (y3)^{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} that is 1 + a^{2} + b^{2} = 18 Since P is on the circle, a^{2} + b^{2}  6a 6b + 17 = 0 Substituting from the equation above gives 6a + 6b = 34 or b = ^{17}/_{3}  a If you now substitute this value for b into a^{2} + b^{2}  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} Cheers, 

