How do you determine of the tangent line at the point (4,3) that lies on...

First, confirm that the point (4,3) is on the circle: ${\displaystyle 4^2+3^2=16+9=25}$.
Next, find ${\displaystyle \frac{dy}{dx}}$ by implicitly assuming y is a function of x, using the Chain Rule, and then doing some algebra:
${\displaystyle 2x+2y\cdot \frac{dy}{dx}=0}$ so that ${\displaystyle \frac{dy}{dx}=-\frac{x}{y}}$.
The slope of the tangent line to the circle at the point (x,y)=(4,3) is therefore
${\displaystyle \frac{dy}{dx}=-\frac{4}{3}.}$
This means the equation of the tangent line to the circle at that point can be written as ${\displaystyle y=-\frac{4}{3}(x-4)+3}$, which can also be written as ${\displaystyle y=-\frac{4}{3}x+\frac{25}{3}}$.