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

Jun 14, 2018

${x}^{x} + {y}^{2} + 4 x + 4 y - 9 = 0$

#### Explanation:

eqn. of circle

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

$\left(a , b\right) = \text{ the centre}$

$r = \text{ the radius}$

centre is the midpoint of the diameter

$\left(a , b\right) = \left(\frac{4 + 0}{2} , \frac{3 + 1}{2}\right)$

$\left(a , b\right) = \left(2 , 2\right)$

$r = \sqrt{{\left(2 - 0\right)}^{2} + {\left(2 - 1\right)}^{2}}$

$r = \sqrt{{4}^{2} + {1}^{2}} = \sqrt{17}$

eqn.

${\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = 17$

${x}^{2} + {y}^{2} - 4 x - 4 y + 8 = 17$

${x}^{x} + {y}^{2} - 4 x - 4 y - 9 = 0$