Circle & Line of Tangency Question

It's been too long since I've done this stuff to know how to get past this point, but I'm thinking you get your points of tangency based on the slope of the linear equation. The slope is 4/3, and there are only two points on your circle where a tangent has a slope of 4/3.

Figure out where those are and you're home free.

Are you in trig or calc?

My school combines all different types of Math into three levels, and I'm in the third level.

How do I "figure out where those are?"

Anyone?

Thanks!

If you differentiate the equation of the circle (I don't remember how to do this) and set dy/dx=4/3, you'll know where the circle has tangents with a slope of 4/3. (Ring any bells?)

There may be a way to do it with trig alone... Perhaps figure out where a line drawn through the circle along its diameter has a slope of -3/4. Your tangents will be perpendicular to that line, and will touch the circle at the same points where this line intersects it.

Both are viable strategies, I think.

Two equations in two unknowns:

x^2 -6x + y^2 + 2y + 1 = 0
3y - 4x = d

Now, solving the second equation for y:

3y = d + 4x
y = d/3 + (4/3)x

Substituting this into the first equation:

x^2 -6x + [d/3 + (4/3)x]^2 + 2[d/3 + (4/3)x] + 1 = 0

Multiplying through by 9 to at least be dealing with whole numbers:

9x^2 -54x + [d + 4x]^2 + 6d + 24x + 1 = 0

9x^2 - 54x + d^2 + 8dx + 16x^2 + 6d + 24x + 1 = 0

25x^2 + (8d - 30)x + d^2 + 6d + 1 = 0

I hope I haven't made any arithmetic errors in getting to this point. This is very messy, but now you can solve the quadratic equation in x, and then use it to obtain a value for y.

Good God, man, I think I'd stick with the geometry of it. Where on a circle does a tangent have the slope 4/3? -- then work from there.