How do you find the center and radius of #(X-5)^2 + Y^2 = 1/16#?

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Answer 1 · 2016-09-17T11:52:14

Centre at #(5,0)# Radius = #1/4#

The equation of a circle with centre at point #(a,b)# and radius #r# is:

#(x-a)^2+(y-b)^2 = r^2#

Here we are given the equation: #(x-5)^2+y^2 = 1/16#
Hence, in this example: #a=5#, #b=0# and #r^2=1/16#

Therefore the centre of the circle is at #(5,0)#

The radius is #1/sqrt16 = 1/4# (Since the radius must be positive)

This can be seen on the graph of the positive portion of the circle below.

graph{(1/16 - (x-5)^2)^(1/2) [4.6156, 5.355, -0.057, 0.3128]}