Question

the equation of the tangent at ( 0 , 2) to the circle with equation 
(x + 2)2 + (y + 1)2 = 13
​

Answer 1

(x + 2)^2 + (y + 1)^2 = 13

Center is at (-2, -1) the slope of the radius from the center to point (0, 2) is:

slope = (2 -(-1))/(0 -(-2)) = 3/2

the radius is perpendicular to the tangent line at the point of tangency and the slopes of the perpendicular lines are negative reciprocal so:

slope of the tangent line = -2/3

the equation of tangent at (0, 2) is:

y - 2 = -2/3(x - 0)

y = (-2/3)x + 2 => answer in slope - intercept form

3y = -2x + 6

2x + 3y = 6 => answer in standard form

Answer 2

Question

the equation of the tangent at ( 0 , 2) to the circle with equation 

(x + 2)2 + (y + 1)2 = 13

ANSWER

Use implicit differentiation to take the derivative of the equation for the circle:

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

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

Now plug in (6,6) for x and y in the dy/dx equation to get

dy/dx= -3/4

This is the slope of the tangent line at (6,6). Now use point slope form to get the equation for the line, using the slope -3/4 and the point (6,6)

y-6=(-3/4)(x-6)

y=(-3/4)x+21/2