### Instruction: What is the slope of the line that is tangent to a circle at point (5,5) if the center of the circle is (3,2)? Express your answer as a common fraction.### Response: Let's start by finding the equation of the circle.
The equation of the circle is $(x-3)^2+(y-2)^2=r^2$.
That's right. Now let's plug in the point (5,5) and see what we get.
$(5-3)^2+(5-2)^2=r^2$ which simplifies to $4+9=r^2$ so $r^2=13$.
Now we can write the equation of the circle as $(x-3)^2+(y-2)^2=13$.
Now let's find the derivative of the equation.
The derivative of $(x-3)^2+(y-2)^2=13$ is $2(x-3)+2(y-2)\cdot\frac{dy}{dx}=0$.
We can simplify that to get $(x-3)+(y-2)\cdot\frac{dy}{dx}=0$.
Now let's plug in the point (5,5) and find $\frac{dy}{dx}$.
Plugging in the point (5,5) into the equation, we get $(5-3)+(5-2)\cdot\frac{dy}{dx}=0$.
This simplifies to $2+3\cdot\frac{dy}{dx}=0$.
So, $\frac{dy}{dx}=\frac{-2}{3}$.