Question

which equations represent circles that have a diameter of 12 units and a center that lies on the y - axis? select two options.
$x^{2}+(y - 3)^{2}=36$
$x^{2}+(y - 5)^{2}=6$
$(x - 4)^{2}+y^{2}=36$
$(x + 6)^{2}+y^{2}=144$
$x^{2}+(y + 8)^{2}=36$

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. Given diameter \(d = 12\), then radius \(r=\frac{d}{2}=6\), so \(r^2 = 36\).

Step2: Check the center location

For the center to lie on the \(y\) - axis, \(h = 0\).

• For the equation \(x^{2}+(y - 3)^{2}=36\), \(h = 0,k = 3,r^2=36\).
• For the equation \(x^{2}+(y - 5)^{2}=6\), \(r^2=6

eq36\).

• For the equation \((x - 4)^{2}+y^{2}=36\), \(h = 4

eq0\).

• For the equation \((x + 6)^{2}+y^{2}=144\), \(r^2=144

eq36\).

• For the equation \(x^{2}+(y + 8)^{2}=36\), \(h = 0,k=-8,r^2 = 36\).

Answer:

\(x^{2}+(y - 3)^{2}=36\), \(x^{2}+(y + 8)^{2}=36\)