Answer
1) The point $(3,-4)$ is on the curve.
2) The tangent line to the curve at $(3,-4)$ is $$y=\frac{3}{4}x-\frac{25}{4}$$
3) The tangent line to the curve at $(3,-4)$ is $$y=-\frac{4}{3}x$$
Work Step by Step
$$x^2+y^2=25$$
1) Verify that the point $(3,-4)$ is on the curve
We substitute the coordinates of the point $(3,-4)$ into the formula of the function to see if it equals $25$ or not:
$$3^2+(-4)^2=9+16=25$$
Since it equals $25$, the point is on the curve.
2) Find the tangent at $(3,-4)$:
- Find the derivative of the function with implicit differentiation:
$$2x+2yy'=0$$ $$x+yy'=0$$ $$yy'=-x$$ $$y'=-\frac{x}{y}$$
- The slope of the tangent line at $(3,-4)$ is $$y'=-\frac{3}{-4}=\frac{3}{4}$$
- The tangent line to the given curve at $(3,-4)$ is
$$y+4=\frac{3}{4}(x-3)$$ $$y+4=\frac{3}{4}x-\frac{9}{4}$$ $$y=\frac{3}{4}x-\frac{25}{4}$$
3) Find the normal line at $(3,-4)$:
The normal line at $(3,-4)$ would be perpendicular with the tangent line at $(3,-4)$, so the product of the slopes of these two lines equal $-1$.
So if we call the slope of the normal line at $(3,-4)$ $k$, we would have
$$k\times\frac{3}{4}=-1$$ $$k=-\frac{4}{3}$$
- The normal line to the given curve at $(3,-4)$ therefore is
$$y+4=-\frac{4}{3}(x-3)$$ $$y+4=-\frac{4}{3}x+4$$ $$y=-\frac{4}{3}x$$
