Answer
The tangent line at (4,3) equals y=-$\frac{4}{3}(x-4)+3$
The normal line at (4,3) equals y=$\frac{3}{4}(x-4)+3$
The tangent line at (-3,4) equals y= $\frac{3}{4}(x+3)+4$
The normal line a (-3,4) equals y= -$\frac{4}{3}(x+3)+4$
Work Step by Step
1. First find the derivate of $x^{2}+y^{2}=25$ by using implicit differentiation.
(d/dx) [ x^{2}+y^{2}=25 ]
2x+ 2y(dy/dx)= 0
Solve for (dy/dx)
dy/dx= (-2x)/(2y)
Simply (2 cancels out)
dy/dx= -x/y
2. Plug in the given points to find the slope of the two tangent lines (since the derivate is the slope #Step 1)
@ (4,3) = -4/3
@ (-3,4) = [-(-3)]/4 =3/4
3. Make the equation of the tangent lines using the point-slope formula
at (4,3) equals y=-$\frac{4}{3}$(x-4)+3
at (-3,4) equals y= $\frac{3}{4}$(x+3)+4
4. Reciprocate (flip the numerator and denominator and change the sign) the slope (of the tangent line) to find the normal slope, then plug it into the point-slope formula.
