Question

The circle passing through the point (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:

By symmetry, if a circle has its center at \(x = 3\) and passes through \((1, -2)\), it must also pass through the point reflected across the vertical axis of the center. Since \(1\) is \(2\) units left of \(3\), the point \(2\) units right of \(3\) (which is \(5\)) with the same y-coordinate \((-2)\) will also lie on the circle.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
When a circle touches the x-axis at a point \((x_1, 0)\), its center must lie on the vertical line \(x = x_1\). Furthermore, the radius of the circle is equal to the absolute value of the y-coordinate of the center.

Step 2: Key Formula or Approach:
The equation of a circle touching the x-axis at \((h, 0)\) is: \[ (x - h)^2 + (y - k)^2 = k^2 \]

Step 3: Detailed Explanation:
1. Since the circle touches the x-axis at \((3, 0)\), its equation is \((x - 3)^2 + (y - k)^2 = k^2\).
2. The circle passes through \((1, -2)\). Substitute these coordinates into the equation: \[ (1 - 3)^2 + (-2 - k)^2 = k^2 \] \[ (-2)^2 + (4 + 4k + k^2) = k^2 \] \[ 4 + 4 + 4k + k^2 = k^2 \]
3. Solve for \(k\): \[ 8 + 4k = 0 \implies 4k = -8 \implies k = -2 \]
4. The equation of the circle is \((x - 3)^2 + (y + 2)^2 = (-2)^2 = 4\).
5. Test the options. For \((5, -2)\): \[ (5 - 3)^2 + (-2 + 2)^2 = 2^2 + 0^2 = 4 \] This satisfies the equation.

Step 4: Final Answer
The circle also passes through the point (5, -2).