Question

an equation of the circle whose center is (-3, 1) and whose radius is 8 is
○ $(x + 3)^2 + (y - 1)^2 = 8$
○ $(x - 3)^2 + (y + 1)^2 = 64$
○ $(x - 3)^2 + (y + 1)^2 = 8$
○ $(x + 3)^2 + (y - 1)^2 = 64$
question 9
1 pts
circle o is inscribed in square efgh, as shown below.
e
f

Explanation:

Step1: Recall the standard circle equation

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

Step2: Identify the center and radius values

Given the center is \((-3, 1)\), so \(h = -3\) and \(k = 1\). The radius \(r = 8\), so \(r^2 = 8^2 = 64\).

Step3: Substitute into the standard equation

Substitute \(h = -3\), \(k = 1\), and \(r^2 = 64\) into the standard equation:

• For the \(x\)-term: \((x - (-3))^2=(x + 3)^2\)
• For the \(y\)-term: \((y - 1)^2\)
• The right - hand side: \(64\)

So the equation of the circle is \((x + 3)^2+(y - 1)^2 = 64\).

Answer:

\((x + 3)^2+(y - 1)^2 = 64\) (the fourth option)