Question

which equation represents a circle with a center at $(-3,-5)$ and a radius of 6 units? $(x + 3)^{2}+(y + 5)^{2}=36$ $(x - 3)^{2}+(y - 5)^{2}=6$ $(x - 3)^{2}+(y - 5)^{2}=36$ $(x + 3)^{2}+(y + 5)^{2}=6$

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: Identify \(h\), \(k\), and \(r\)

Given the center \((-3,-5)\), so \(h=-3\), \(k = -5\). The radius \(r = 6\).

Step3: Substitute \(h\), \(k\), and \(r\) into the standard form

Substitute \(h=-3\), \(k=-5\), \(r = 6\) into \((x - h)^2+(y - k)^2=r^2\).
For the \(x\)-part: \(x-(-3)=x + 3\).
For the \(y\)-part: \(y-(-5)=y + 5\).
For the \(r^2\) part: \(r^2=6^2 = 36\).
So the equation is \((x + 3)^2+(y + 5)^2=36\).

Answer:

\((x + 3)^2+(y + 5)^2=36\) (the first option)