# The point (4,7) lies on the circle centered at (-3,-2), how do you find the equation of the circle in standard form?

Feb 1, 2016

${\left(x + 3\right)}^{2} + {\left(y + 2\right)}^{2} = 130$

#### Explanation:

the equation of a circle in standard form is :

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

where (a , b ) is the centre and r , the radius

In this question the centre is given but require to find r

the distance from the centre to a point on the circle is radius.

calculate r using $\textcolor{b l u e}{\text{ distance formula }}$

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

using  (x_1 , y_1 ) = (-3,-2))color(black)(" and") (x_2 , y_2) = (4,7)

then $r = \sqrt{4 - {\left(- 3\right)}^{2} + \left(7 - {\left(- 2\right)}^{2}\right)} = \sqrt{49 + 81} = \sqrt{130}$

circle equation using centre =(a , b ) = (-3 , -2) , r $= \sqrt{130}$

$\Rightarrow {\left(x + 3\right)}^{2} + {\left(y + 2\right)}^{2} = 130$