Question

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

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. Given diameter \(d = 12\), then radius \(r=\frac{d}{2}=6\). Also, since the center lies on the \(y\) - axis, \(h = 0\).

Step2: Check each equation

• For \((x + 6)^2+y^2=144\), here \(r^2 = 144\), so \(r = 12\) (not valid as radius should be \(6\)).
• For \(x^2+(y + 8)^2=36\), \(r^2=36\) (\(r = 6\)), and \(h = 0\) (center \((0,-8)\) lies on \(y\) - axis).
• For \(x^2+(y - 3)^2=36\), \(r^2=36\) (\(r = 6\)), and \(h = 0\) (center \((0,3)\) lies on \(y\) - axis).
• For \(x^2+(y - 5)^2=6\), \(r^2=6\) (\(r=\sqrt{6}

eq6\)).

• For \((x - 4)^2+y^2=36\), \(h = 4

eq0\) (center \((4,0)\) does not lie on \(y\) - axis).

Answer:

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