Answer
(a) $y=\frac{3}{4}x-\frac{25}{4}$
(b) $y=-\frac{4}{3}x$
Work Step by Step
To verify that the given point is on the curve, plug-in the coordinates to the left side of the equation: $LHS=(3)^2+(4)^2=9+16=25=RHS$
(a) Given the equation $x^2+y^2=25$, differentiate both sides with respect to $x$: $2x+2yy'=0$, which gives $y'=-\frac{x}{y}$
Thus the slope of the tangent line at the given point is $m=y'=-\frac{x}{y}=-\frac{3}{-4}=\frac{3}{4}$ and the equation for the tangent line is $y+4=\frac{3}{4}(x-3)$ or $y=\frac{3}{4}x-\frac{25}{4}$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-\frac{4}{3}$, and the line equation is $y+4=-\frac{4}{3}(x-3)$ or $y=-\frac{4}{3}x$
