# Find the equation of the circle in general forn if the diameter is 12 and the center is 8,1? Thabk you for the answer

Jun 20, 2018

${\left(x - 8\right)}^{2} + {\left(y - 1\right)}^{2} = {6}^{2}$

#### Explanation:

If the diameter is $12$ then the radius is $\frac{12}{2} = 6$

The general equation of a circle with center $\left(a , b\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

Jun 25, 2018

${\left(x - 8\right)}^{2} + {\left(y - 1\right)}^{2} = {6}^{2}$

#### Explanation:

The equation of a circle is basically using Pythagoras ${a}^{2} + {b}^{2} = {c}^{2}$

It is just that we are using the x-axis value as say $a$, the y-axis value as say $b$ and the radius of the circle as $c$

If the centre is at the origin we have ${x}^{2} + {y}^{2} = {r}^{2}$

This works very well when centred at at the origin but what if it is elsewhere?

We 'hypothetically' take it back to the origin

So if the centre is at 8 on the x-axis we write $\left(x - 8\right)$
So if the centre is at 1 on the y-axis we write $\left(y - 1\right)$

The length of the radius is unchanged giving:

${\left(x - 8\right)}^{2} + {\left(y - 1\right)}^{2} = {r}^{2} = {\left(\frac{12}{2}\right)}^{2}$