Calculus problem!!!!!

Find an equation of the tangent line to the graph at the given point.

(x+2)^2 + (y-3)^2 = 37 (-8,4)


y = ???

Take the derivative to find the slope of the tangent line.

2(x+2) + 2(y-3) dy/dx = 0

2(-8+2) + 2(4-3) dy/dx = 0

-12 + 2 dy/dx = 0

dy/dx = slope of tangent line = 6

y-y1=m(x-x1)

y-4 = 6(x-(-8))

y = 6x + 52

you need to differentiate to find slope:

(x+2)^2 + (y-3)^2 = 37
x^2 + 4x + 4 + y^2 - 6y + 9 = 37 ! expand
2x + 4 + 2y(dy/dx) - 6 = 0 ! differentiate
dy/dx = (2 - 2x) / (2y) ! solve for the derivative in terms of x and y
dy/dx = (1 - x) / y
dy/dx = 9 / 4 ! plug in

Now use that slop to find the x intercept (assuming you want it in the form y = mx + b)

y - 4 = (x - -8) * (9/4)
y = (9/4) * x + 22