Question

Find the center and radius of (x + 9)2 + (y + 6)2 = 9.
(–9, –6); 3

(–6, –9); 9

(9, 6); 3

(–6, 9); 3

Answer 1

Alright, what do you know about circles?

Answer 2

The equation of a circle at origin is

x^2 + y^2 = r^2.

Part of that still applies here. r^2.

Since it's asking for the radius, we find the square root of 9. Mind telling me what that is?

Answer 3

(x-a)^2+(y-b)^2 = r^2

Answer 4

$\left(x+9\right)^2+\left(y+6\right)^2=9$(x+9)2+(y+6)2=9




Hide Definition
$\mathrm{Circle\:Equation}$Circle Equation

$\left(x−a\right)^2+\left(y−b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right)$(x−a)2+(y−b)2=r2 is the circle equation with a radius r, centered at (a, b)



$\mathrm{Rewrite}\:\left(x+9\right)^2+\left(y+6\right)^2=9\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}$Rewrite (x+9)2+(y+6)2=9 in the form of the standard circle equation

$\left(x-\left(-9\right)\right)^2+\left(y-\left(-6\right)\right)^2=3^2$(x−(−9))2+(y−(−6))2=32



$\mathrm{Therefore\:the\:circle\:properties\:are:}$Therefore the circle properties are:

$\left(a,\:b\right)=\left(-9,\:-6\right),\:r=3$(a, b)=(−9, −6), r=3

Answer 5

there u go

Answer 6

\[$\left(x+9\right)^2+\left(y+6\right)^2=9$(x+9)2+(y+6)2=9 Hide Definition $\mathrm{Circle\:Equation}$Circle Equation $\left(x−a\right)^2+\left(y−b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right)$(x−a)2+(y−b)2=r2 is the circle equation with a radius r, centered at (a, b) $\mathrm{Rewrite}\:\left(x+9\right)^2+\left(y+6\right)^2=9\:\mathrm{\in\:the\:form\:of\:the\:standard\:circle\:equation}$Rewrite (x+9)2+(y+6)2=9 \in the form of the standard circle equation $\left(x-\left(-9\right)\right)^2+\left(y-\left(-6\right)\right)^2=3^2$(x−(−9))2+(y−(−6))2=32 $\mathrm{Therefore\:the\:circle\:properties\:are:}$Therefore the circle properties are: $\left(a,\:b\right)=\left(-9,\:-6\right),\:r=3$(a, b)=(−9, −6), r=3\]