# How do you write the equation for a circle with points (0, 2a) (2b, 0) as ends of a diameter?

Jan 22, 2017

The x coordinate goes from 0 to 2b so the center is at b. Likewise, the y coordinate goes from 2a to 0 so the center is at a.
Use the equation (x-h)^2+(y-k)^2=r^2; h=b,k=a and one of the points.

#### Explanation:

Use the equation
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where $h = b \mathmr{and} k = a$

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

Let's use the point $\left(0 , 2 a\right)$, to find the value of r:

${\left(0 - b\right)}^{2} + {\left(2 a - a\right)}^{2} = {r}^{2}$

${r}^{2} = {a}^{2} + {b}^{2}$

$r = \sqrt{{a}^{2} + {b}^{2}}$

Here is the equation of the circle:

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