Question

Let a circle pass through the point \( (1, 2) \) and touch the \( x \)-axis at \( (3, 0) \). Then the equation of the circle is:

• A. \( (x - 3)^2 + (y - 4)^2 = 16 \)
• B. \( (x - 3)^2 + y^2 = 16 \)
• C. \( (x - 3)^2 + (y - 2)^2 = 4 \)
• D. \( (x - 3)^2 + (y - 1)^2 = 1 \)

Hint:

When a circle touches the \( x \)-axis at a point, the \( x \)-coordinate of the center matches that point, and the radius equals the \( y \)-coordinate of the center.

Answer 1

The Correct Option is C

- The general equation of a circle is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius.
- The circle touches the \( x \)-axis at \( (3, 0) \), so the center is at \( (3, k) \), and the radius \( r = k \) (distance to the \( x \)-axis). The equation is: \[ (x - 3)^2 + (y - k)^2 = k^2 \]
- The circle passes through \( (1, 2) \). Substitute \( (x, y) = (1, 2) \): \[ (1 - 3)^2 + (2 - k)^2 = k^2 \implies 4 + (2 - k)^2 = k^2 \implies 4 + 4 - 4k + k^2 = k^2 \implies 8 - 4k = 0 \implies k = 2 \]
- Thus, \( r = k = 2 \), and the equation is: \[ (x - 3)^2 + (y - 2)^2 = 4 \]