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Find all circles that pass through the point (3, -6) and are tangent to both coordinate axes.

1. x² + y² - 6x + 6y + 9 = 0
2. x² + y² + 6x - 6y + 9 = 0
3. x² + y² + 30x - 30y + 225 = 0
4. x² + y² - 30x + 30y + 225 = 0

Correct answer: x² + y² - 6x + 6y + 9 = 0

Solution

Since the circle must pass through (3, -6) which is in the fourth quadrant, the relevant center forms are (r, -r). Solving (3-r)² + (-6+r)² = r² gives r = 3 or r = 15, yielding two valid circles.

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