1. ## tangent line

"Sketch the circles x²+y²=1 and y²+(x-3)²=4. There is a line with positive slope that is tangent to both circles. Find the points at which the tangent line touches each circle."

solve using implicit differentiation

I found the answer online and it is (-1/3, (2√2)/3) and (7/3, (4√2)/3), but i dont know how he got it.

i know the slopes of the circles are -x/y and (3-x)/y
can this be a system of equations?

2. Originally Posted by zintore
"Sketch the circles x²+y²=1 and y²+(x-3)²=4. There is a line with positive slope that is tangent to both circles. Find the points at which the tangent line touches each circle."

solve using implicit differentiation

I found the answer online and it is (-1/3, (2√2)/3) and (7/3, (4√2)/3), but i dont know how he got it.

i know the slopes of the circles are -x/y and (3-x)/y
can this be a system of equations?
No, they are the gradients at each point on the respective circles.

There's very likely some circle theorems that can be used to solve this problem easily (the two circles are tangent at (1, 0) and you know the radii of each circle) but in the meantime your stuck with this approach:

Let the gradient of the tangent be m.

Let the point on the circle $\displaystyle (x-3)^2 + y^2 = 4$ be (a, b). Then:

$\displaystyle (a-3)^2 + b^2 = 4$ ..... (1)

$\displaystyle \frac{3-a}{b} = m$ .... (2)

Let the point on the circle $\displaystyle x^2 + y^2 = 1$ be (e, f). Then:

$\displaystyle e^2 + f^2 = 1$ ..... (3)

$\displaystyle - \frac{e}{f} = m$ .... (4)

Also, $\displaystyle \frac{b-f}{a-e} = m$ .... (5)

Solve equations (1) to (5) simultaneously for a, b, e, f. I'd suggest eliminating m and solve the resulting four equations in the unknowns a,,b, e, f.