given the circle whose equation is (x + 4)^2 + (y - 2)^2)^2 = 136, write the equation of the tangent to the circle at point (2,12).

Question

PLEASEREE show all work

Answer 1

This can be answered without using calculus.

You need to know that when a radius of a circle intersects with a tangent line they meet at a 90 degree angle (they are perpendicular to each other). In this problem the endpoints of the radius are (-4,7) and (2,12). The slope of the radius is (12-2)/(2+4) = 10/6. Therefore the slope of the tangent line must be -6/10 or -3/5 in order for the tangent line to be perpendicular to the radius.

The next step is to find the equation of the line using the point-slope form, which in this case would be

y-12 = (-3/5)(x-2).

Answer 2

The radius is perpendicular to the tangent at that point (2,12) on the circle.

The endpoints of the radius are the center of the circle (-4,2) and the point on the circle (2,12)

The slope of the radius is (12-2)/(2--4) = 5/3.

The slope of the perpendicular tangent will be -3/5

In point slope form the equation of the tangent line is:

y - 12 = -3/5 (x - 2)

using algebra you can get it into y = mx + b form.

Answer 3

(x+4)^2 + (y-2)^2 = 136

point (2, 12)

it is a point on the circle as (6^2 + 10^2) = 136. That's worth checking as some problems get miscopied.

it helps to draw a sketch, at least a rough graph, plotting the point and circle

It give you an idea of the slope of the tangent line. Circle's center is (-4, 2) with radius = sqr136 = about 11.7

slope is downward sloping <0, not -1, but close to it

(x+4)^2 + (y-2)^2 = 136

expand

x^2 + 8x + 16 + y^2 -4y + 4 = 136

take the derivative

2x + 8 +2yy' -4y' =0

solve for y'

y'(2y-4) = -2x-8

y' = (-2x-8)/(2y-4)

plug in the point (2,12)

y' = (-4-8)/(24-4) = -12/20 =-3/5 = the slope of the tangent line to the cirvle, at the point (2,12)

the tangent line's equation is

-3/5 = (y-12)/(x-2) in point slope form

cross multiply

5y -60 = -3x +6

3x+5y = 66 in standard form

Answer 4

(x+4)² + (y-2)² = 136

Find dy/dx implicitly: 2(x+4) + 2(y-2)(dy/dx) = 0

At the point (2,12), we have 12 + 20(dy/dx) = 0

So, dy/dx = -12/20 = -3/5 = slope of tangent line at (2,12).

Equation of tangent line: y - 12 = -(3/5)(x - 2)