# How do you write the standard from of the equation of the circle given center:(2,-6) radius: 2?

Jun 3, 2016

$\textcolor{g r e e n}{{\left(x - 2\right)}^{2} + {\left(y + 6\right)}^{2} = {2}^{2}}$

See explanation as to why it is in this form.

#### Explanation:

The standardised formula for a circle with its centre at the origin is:

${x}^{2} + {y}^{2} = {r}^{2}$

This is derived from Pythagoras rule about a 'right triangle'. You know the one; ${a}^{2} + {b}^{2} = {c}^{2}$

However, in this case the centre is not at the origin. It is at

$\left(x , y\right) \to \left(2 , - 6\right)$

Consequently we mathematically move all the points as if the centre of the circle was at the origin so that $\left(x , y\right) \to \left(0 , 0\right)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Moving the } x ' s}$
${x}_{\text{moved")=x_("given}} - 2$

so each generic point we have $x - 2$

$\textcolor{b r o w n}{\text{Moving the } y ' s}$
${y}_{\text{moved")=y_("given}} + 6$

so each generic point we have $y + 6$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{g r o w n}{\implies {\left({x}_{\text{generic"))^2+(y_("generic}}\right)}^{2} = {r}^{2}}$

Becomes:

$\textcolor{g r e e n}{{\left(x - 2\right)}^{2} + {\left(y + 6\right)}^{2} = {2}^{2}}$ 