# How do we find the standard equation of the circle with endpoints of a diameter (1/2, 4) (3/2, -1)?

Jun 15, 2017

${\left(x - 1\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = {2.55}^{2}$

#### Explanation:

The standard equation of a circle takes the form:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {c}^{2}$ where $c$ is the radius

We need to find the radius, $r$, and the location of the centre, the point $\left(a , b\right)$

The length of the diameter is given by:

$d = 2 r = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$d = \sqrt{{\left(\frac{3}{2} - \frac{1}{2}\right)}^{2} + {\left(- 1 - 4\right)}^{2}}$
$= \sqrt{{1}^{2} + {\left(- 5\right)}^{2}} = \sqrt{1 + 25} = 5.1$ units

The radius will be half the diameter, $2.55$ units

We can then use the midpoint formula to find the coordinates of the centre. The midpoint formula simply averages the values of the endpoints:

$\left(a , b\right) = \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right) = \left(\frac{\frac{1}{2} + \frac{3}{2}}{2} , \frac{4 - 1}{2}\right) = \left(1 , \frac{3}{2}\right)$

Now we have all the information we need to write the equation for the circle:

${\left(x - 1\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = {2.55}^{2}$

Jun 15, 2017

${\left(x - 1\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = \frac{13}{2}$

#### Explanation:

$\text{the standard form of the equation of a circle is }$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where ( a , b ) are the coordinates of the centre and r the radius.

$\text{the centre will be at the midpoint of the diameter}$

$\text{midpoint "=[1/2(1/2+3/2),1/2(4-1)]}$

$\Rightarrow \text{centre } = \left(1 , \frac{3}{2}\right) \leftarrow \left(a , b\right)$

$\text{the radius is the distance from the centre to either of }$
$\text{the endpoints}$

$\text{to calculate r use the "color(blue)"distance formula}$

$\text{points are " (1,3/2)" and } \left(\frac{3}{2} , - 1\right)$

$r = \sqrt{{\left(\frac{3}{2} - 1\right)}^{2} + {\left(- 1 - \frac{3}{2}\right)}^{2}}$

$\textcolor{w h i t e}{r} = \sqrt{\frac{1}{4} + \frac{25}{4}} = \sqrt{\frac{13}{2}}$

$\Rightarrow {\left(x - 1\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = {\left(\sqrt{\frac{13}{2}}\right)}^{2}$

$\Rightarrow {\left(x - 1\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = \frac{13}{2} \text{ is the equation}$