Implicit Differentiation
Because this question was posted under Implicit Differentiation, we'll do it that way.
We want #dy/dx# given that #x# and #y# are related by:
#x^2 + y^2 = 12#
Differentiate both sides with respect to #x#. Remember that we are considering #y# to be some function of #x# that we haven't found. It often helps to think of #y# as "some stuff in parentheses".
When we differentiate "some stuff in parentheses" squared, we gat #2# times that stuff times the derivative of the stuff (because of the chain rule).
Also: remember that #d/dx# of some expression means the derivative (with respect to x) of that expression.
o, (in moderate detail)
#x^2 + y^2 = 12#
#d/dx(x^2 + y^2) = d/dx(12)#
#d/dx(x^2) + d/dx(y^2) = d/dx(12)#
#2x + underbrace(2y dy/dx)_"used chain rule" = 0#
#dy/dx = -x/y#
At #(sqrt3,3)#, the slope is #m = -sqrt3/3#
The equation of the line through that point with that slope is:
#y=-sqrt3/3x+4#
Notes
(1) because we see that the #y# coordinate of the point of interest is positive, we know that the point lies on the upper half of the circle:
#x^2 + y^2 = 12#
Solving for #y# on that semicircle gets us #y = sqrt(12-x^2)# and we do not need implicit differentiation.
(2) It is a circle centered at the origin, The tangent at #(sqrt3,3)# is perpendicular to the radius connecting #(0,0)# and #(sqrt3, 3)#. Find the slope of the radius and then the slope of a perpendicular to the radius. No need for calculus.
