# How do you find general form of circle with endpoints of a diameter at the points (4, 3) and (0, 1)?

Dec 31, 2015

General Form: ${\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = 5$
or ${x}^{2} - 4 x + {y}^{2} - 4 y + 3 = 0$

#### Explanation:

The center of the circle is the middle point of the diameter
${x}_{c} = \frac{4 + 0}{2} = 2$ and ${y}_{c} = \frac{3 + 1}{2}$. Then C(2,2).

The distance between the center of the circle and to a endpoint of a diameter is equal to the radius:
$R = \sqrt{{\left(4 - 2\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{4 + 1} = \sqrt{5}$

The formula of the general form of the circle is
${\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {R}^{2}$

Then, in case, the equation of the circle is
${\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = 5$