4. The normal to the cirlcle x^2 - 4x + y^2 = 9 at (5,2)
Let's complete the square to find the center
x^2 -4x + 4 + y^2 = 4 + 9
(x - 2)^2 + (y - 0)^2 = 13
The center is (2, 0)
The slope between this point and ( 5,2) is
[ 2- 0 ] / [ 5 - 2 ] = 2/3 = slope of the normal line
The equation of the normal line at (5,2) is
y = (2/3) (x - 5) + 2
y = (2/3)x - 10/3 + 2
y= (2/3)x - 4/3
Here's the graph : https://www.desmos.com/calculator/u8n7sc01uh
