If a circle has a center at the point (3, 4) and goes through the point (6, 8), what is the radius?

We know that the standard form of a circle is given as:

(x – h)^2 + (y – k)^2 = r^2

where the variables are,

h and k are the center points of the circle

x and y are points on the circle

r is the radius

 

In this case, we are given that:

(h, k) = (3, 4)

(x, y) = (6, 8)

 

Therefore substituting these values into the equation and calculating for radius, r:

(6 – 3)^2 + (8 – 4)^2 = r^2

r^2 = 3^2 + 4^2

r^2 = 9 + 16

r^2 = 25

r = 5

Therefore the radius of the circle is 5 units.

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