Use calculus to show that the closest point on the graph of x^2 + y^2 = 5^2 to the point (6,8) is point (3,4)

Answer 1

x^2 + y^2 = 5^2 is a circle with center at the origin and radius = 5

the point (6,8) is closest to the circle, at the point (3,4) where the tangent line has slope which is the negative inverse of the line from (3,4) to (6,8). the line's slope = (8-4)/(6-3) = 4/3

the tangent line has slope dy/dx=y'

y^2 = 25-x^2

2y'y = -2x

y'= -x/y

which at (3,4) = -3/4 which is the negative inverse of 4/3

it might help to draw a sketch, a rough graph, plot the points

draw the circle and the lines