# What is the equation of a circle with the diameter being a line joining the points (3, 2) and (4, 6)

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The equation of a circle has to be determined where the diameter is a line segment joining the points (3, 2) and (4, 6).

First, determine the center of the circle. This is the mid point of the line segment. The coordinates of the point are `((3 +4)/2, (2+6)/2) = (3.5, 4)`

The radius of the circle is half the distance between the two points. This is equal to `(1/2)*sqrt((4-3)^2+(6-2)^2)` = `(1/2)*sqrt(1+16)` = `(sqrt 17)/2`

The equation of a circle with radius `(sqrt 17)/2` and center (3.5, 4) is `(x - 3.5)^2 + (y-4)^2 = ((sqrt 17)/2)^2`

=> `(x - 3.5)^2 + (y-4)^2 = 17/4`

**The equation of the required circle is `(x - 3.5)^2 + (y-4)^2 = 17/4` **

The equation of a circle with center (a, b) and radius r is `(x - a)^2 + (y - b)^2 = r^2`

If the diameter of a circle connects the points (3, 2) and (4, 6), the center is the mid point between the two points and the radius is half the distance between them.

The center of the circle is: `((3+4)/2, (2+6)/2) = (3.5, 4)`

The radius of the circle is: `(1/2)*(sqrt((3 - 4)^2 + (2 - 6)^2)`

= `(1/2)*(sqrt(1 + 16)`

= `sqrt17/2`

The equation of the circle is `(x - 3.5)^2 + (y - 4)^2 = 17/4`