A circle is tangent to the x-axis at (5, 0) and passes through the point (8, 6). What is its radius? - IQnection

Understanding the Context

Why This Problem Is Gaining Attention

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Key Insights

How It Works: The Mathematics Behind the Tangent Circle

When a circle is tangent to the x-axis at point (5, 0), its center lies directly above or below this point at a distance equal to the radius, because tangency guarantees a single point of contact. Since the circle lies above the x-axis (evidenced by passing through (8, 6), which is above the axis), the center must be at (5, r), where r is the radius.

This center point forms a right triangle with both (5, 0) and (8, 6). The horizontal leg spans 8 – 5 = 3 units, and the vertical leg spans 6 – r units. By the Pythagorean theorem:

[ r^2 = 3^2 + (6 - r)^2 ]

Final Thoughts

Expanding the right side:

[ r^2 = 9 + (36 - 12r + r^2) ]

Simplifying:

[ r^2 = r^2 - 12r + 45 ]

Subtract ( r^2 ) from both sides:

[ 0 = -12r + 45 ]