Question

which of the following equations represents a circle in the xy - plane that is tangent to the y - axis at exactly one point?
a) ((x - 12)^2 + (y - 12)^2 = 16)
b) ((x - 12)^2 + (y - 6)^2 = 36)
c) ((x - 6)^2 + (y - 8)^2 = 36)
d) (x^2 + (y - 8)^2 = 36)

Explanation:

Step1: Recall the circle 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. A circle tangent to the \(y\)-axis means the distance from the center \((h, k)\) to the \(y\)-axis (which is \(|h|\)) equals the radius \(r\).

Step2: Analyze each option

• Option A: Center \((12, 12)\), radius \(r = \sqrt{16}=4\). Distance from center to \(y\)-axis is \(|12| = 12

eq4\), not tangent.

• Option B: Center \((12, 6)\), radius \(r=\sqrt{36} = 6\). Distance from center to \(y\)-axis is \(|12| = 12

eq6\), not tangent.

• Option C: Center \((6, 8)\), radius \(r=\sqrt{36}=6\). Distance from center to \(y\)-axis is \(|6| = 6\), which equals the radius. So this circle is tangent to the \(y\)-axis.
• Option D: Center \((0, 8)\), radius \(r=\sqrt{36}=6\). Distance from center to \(y\)-axis is \(|0| = 0

eq6\) (it's on the \(y\)-axis, not tangent at one point).

Answer:

C. \((x - 6)^2 + (y - 8)^2 = 36\)