<p>The circle passing through \((1, -2)\) and touching the axis of x at \((3, 0)\) also passes through the point:</p>
Step-by-Step Solution
Key Concept: When a circle touches the x-axis at a point, its center lies on the perpendicular to the x-axis at that point, making the center's x-coordinate equal to the tangent point's x-coordinate.
<p>The circle passes through \((1, -2)\) and touches the x-axis at \((3, 0)\). When a circle touches the x-axis at a point, the center lies on the vertical line through that point. So the center is at \((3, h)\) for some h. Using the condition that the circle passes through \((1, -2)\):</p><p>\[(3-1)^2 + (h-(-2))^2 = h^2\]</p><p>\[4 + (h+2)^2 = h^2\]</p><p>\[4 + h^2 + 4h + 4 = h^2\]</p><p>\[4h = -8\]</p><p>\[h = -2\]</p><p>The center is \((3, -2)\) and radius is 2. The equation is \((x-3)^2 + (y+2)^2 = 4\). Testing \((5, -2)\): \((5-3)^2 + (-2+2)^2 = 4 + 0 = 4\). ✓</p><p>∴ Answer is (d) \((5, -2)\).</p>
Correct Answer: D