Question

what is the standard form of the equation of the circle shown below?
a $(x - 2)^{2}+(y - 2)^{2}=16$ b $(x + 2)^{2}+(y - 2)^{2}=16$
c $(x - 2)^{2}+(y + 2)^{2}=16$ d $(x + 2)^{2}+(y + 2)^{2}=16$

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: Determine the center of the circle

From the graph, the center of the circle is \((- 2,2)\). So \(h=-2\) and \(k = 2\).

Step3: Determine the radius of the circle

Counting the units from the center to a point on the circle (e.g., from \((-2,2)\) to \((2,2)\)), the radius \(r = 4\). Then \(r^2=16\).

Step4: Substitute \(h\), \(k\), and \(r^2\) into the standard - form equation

Substitute \(h=-2\), \(k = 2\), and \(r^2 = 16\) into \((x - h)^2+(y - k)^2=r^2\). We get \((x-(-2))^2+(y - 2)^2=16\), which simplifies to \((x + 2)^2+(y - 2)^2=16\).

Answer:

B. \((x + 2)^2+(y - 2)^2=16\)