find normal line equation of the curve y=sqrt(25-x^2) at the point (3,4)

$$\\y=\sqrt{25-x^2}\\\\
y=(25-x^2)^{0.5}\\\\
\frac{dy}{dx}=0.5(25-x^2)^{-0.5}\times -2x\\\\
\frac{dy}{dx}=-x(25-x^2)^{-0.5}\\\\
when \;\;x=3\\\\
\frac{dy}{dx}=-3(25-3^2)^{-0.5}\\\\
\frac{dy}{dx}=-3(16)^{-0.5}\\\\
\frac{dy}{dx}=-3\times\frac{1}{4}\\\\
\frac{dy}{dx}=-\frac{3}{4}\\\\$$

$$\\$Gradient of the tangent is $ -\frac{3}{4}\\\\
$Gradient of the normal is $ +\frac{4}{3}\\\\\\
\frac{4}{3}=\frac{y-4}{x-3}\\\\
4(x-3)=3(y-4)\\\\
4x-12=3y-12\\\\
$the equation of the normal at (3,4) is $
4x-3y=0$$

y = (25 - x^2)^1/2    taking the derivative, we have

y' = (1/2)(25 -x^2)^(-1/2) (-2x) =  -x / √[25 - x^2]

And the slope of the line when x = 3 = -3/ √[25 - 3^2]  = -3/ √16 = -3/4

And the slope of the normal line will be the negative reciprocal of this = 4/3

So we have

y - 4 = (4/3)(x -3)

y = (4/3)x - 4 + 4

y = (4/3)x       and that's the equation of the normal line to the function at (3, 4)

See the pic here of the function, the tangent line to the function at (3, 4) and the normal line to the function going through the same point

https://www.desmos.com/calculator/ff3alev1sg