Question

The circle passing through \( (1, -2) \) and touching the \(x\)-axis at \( (3, 0) \) also passes through the point:

• A. \( (2, -5) \)
• B. \( (-5, -2) \)
• C. \( (-2, 5) \)
• D. \( (-5, 2) \)
• E. \( (5, -2) \)

Hint:

If a circle touches the \(x\)-axis, its center lies on a vertical line through the point of contact, and radius equals the vertical distance to the axis.

Answer 1

The Correct Option is B

Concept: If a circle touches the \(x\)-axis at a point, then:
• The radius is perpendicular to the \(x\)-axis at the point of contact.
• Hence, the center lies vertically above or below the point of contact.

Step 1: Find the center of the circle. Since the circle touches the \(x\)-axis at \( (3,0) \), the center must lie on the vertical line \( x = 3 \). Let the center be: \[ (3, k) \] Radius: \[ r = |k| \]

Step 2: Use the point \( (1, -2) \). Distance from center to this point equals radius: \[ \sqrt{(1-3)^2 + (-2 - k)^2} = |k| \] Square both sides: \[ (1-3)^2 + (-2 - k)^2 = k^2 \] \[ 4 + (k+2)^2 = k^2 \] \[ 4 + k^2 + 4k + 4 = k^2 \] \[ 8 + 4k = 0 \] \[ k = -2 \] So, center is: \[ (3, -2), r = 2 \]

Step 3: Equation of the circle. \[ (x-3)^2 + (y+2)^2 = 4 \]

Step 4: Check options. Check point \( (-5, -2) \): \[ (-5-3)^2 + (-2+2)^2 = (-8)^2 + 0 = 64 \neq 4 \] Check point \( (5, -2) \): \[ (5-3)^2 + 0 = 4 \checkmark \]