1. <b>Implicit Functions... Help?? </b>

Sketch the circles y^2+x^2=1 and y^2+(x-3)^2=4. There is a line with positive slope thaqt is tangent to both circles. Determine the points at which this tangent line touches each circle.

If you could help it would be greatly appreciated. I'm taking University calculus with only a pre-calculus background, and it's not the easiest thing in the world, let me tell you!

2. We want the equation of the line with positive slope that is tangent two the two circles:

$\displaystyle y=\sqrt{1-x^{2}}, \;\ y=\sqrt{4-(x-3)^{2}}$

If you let $\displaystyle A(a,\sqrt{1-a^{2}})$ be a point on

$\displaystyle y=\sqrt{1-x^{2}}$, then $\displaystyle y'=\frac{-x}{\sqrt{1-x^{2}}}$ and it slope at x=a is $\displaystyle \frac{-a}{\sqrt{1-a^{2}}}$.

Also, let $\displaystyle B(b, \sqrt{4-(b-3)^{2}})$ be a point on

$\displaystyle y=\sqrt{4-(x-3)^{2}}$, then $\displaystyle y'=\frac{3-x}{\sqrt{-x^{2}+6x-5}}$ and its slope at x=b is $\displaystyle \frac{3-b}{\sqrt{-b^{2}+6b-5}}$

These slopes have to be the same, so setting them equal and solving for a we get $\displaystyle a=\frac{b-3}{2}$..........[1]

Just using the slope formula, the slope of AB is:

$\displaystyle \frac{\sqrt{4-(b-3)^{2}}-\sqrt{1-a^{2}}}{b-a}$

This has to be the same as the slope at A:

$\displaystyle \frac{\sqrt{4-(b-3)^{2}}-\sqrt{1-a^{2}}}{b-a}=\frac{-a}{\sqrt{1-a^{2}}}$

Sub in [1] and solve for b and we get:

$\displaystyle b=\frac{7}{3}$

This means that $\displaystyle a=\frac{-1}{3}$

This gives a slope of $\displaystyle \frac{\sqrt{2}}{4}$

So, we have two points and a slope. Almost there.

$\displaystyle A(\frac{-1}{3},\frac{2\sqrt{2}}{3}), \;\ B(\frac{7}{3},\frac{4\sqrt{2}}{3})$

Using these we find we have a line equation of:

$\displaystyle \boxed{y=\frac{\sqrt{2}}{4}x+\frac{3\sqrt{2}}{4}}$

3. Thank-You soooooo much!