A circle centered at (3, 2) is tangent to the line y=(x+1)/3. Find the area of the circle.

Guest May 7, 2019

#1**+1 **

(3,2)

The line can be written as y = (1/3)(x + 1) (1)

We need to find a perpendicular line to this one passing through (3,2)

This line will have a slope of -3.....so we have

y = -3 ( x -3) + 2

y = -3x + 11 (2)

Set (1) and (2) equal to find the x coordinate of the point on the circle where these lines intersect

(1/3)(x + 1) = -3x + 11 multiply through by 3

x + 1 = -9x + 33 rearrange as

10x = 32

x = 32/10 = 16/5

And y = -3(16/5) + 11 = -48/5 + 55/5 = 7/5

So....the intersection is (16/5, 7/5)

And the distance between this point and (3,2) will be the radius of the circle...so....we can find r^2 as

(3 - 16/5)^2 + (2 - 7/5)^2 =

(-1/5)^2 + (3/5)^2 =

1/25 + 9/25 =

10/25 = 2/5 = r^2

So....the area of the circle = pi * r^2 = (2/5) pi units^2

Here's a graph : https://www.desmos.com/calculator/tklpk6vonx

CPhill May 8, 2019