Question

What will be the equation of the circle whose center is \( (1, 2) \) and touches the X-axis?

• A. \( x^2 + y^2 - 2x - 4y + 1 = 0 \)
• B. \( x^2 + y^2 + 2x + 4y + 1 = 0 \)
• C. \( x^2 + y^2 - 2x - 4y - 1 = 0 \)
• D. \( x^2 + y^2 + 2x + 4y - 1 = 0 \)

Hint:

When the circle touches the X-axis, the radius is equal to the y-coordinate of the center.

Answer 1

The Correct Option is A

The equation of a circle with center \( (h, k) \) and radius \( r \) is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] Given the center \( (h, k) = (1, 2) \), we know the circle touches the X-axis. The radius is equal to the y-coordinate of the center, which is \( 2 \). Substitute into the standard circle equation: \[ (x - 1)^2 + (y - 2)^2 = 2^2 \] Expanding and simplifying: \[ (x^2 - 2x + 1) + (y^2 - 4y + 4) = 4 \] \[ x^2 + y^2 - 2x - 4y + 5 = 4 \] \[ x^2 + y^2 - 2x - 4y + 1 = 0 \] Thus, the equation of the circle is \( x^2 + y^2 - 2x - 4y + 1 = 0 \).