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A circle of radius 3 touches the circle x² + y² - 4x - 6y - 12 = 0 internally at the point (-1, -1). Find its equation.
- x² + y² + 2x + 2y - 7 = 0
- x² + y² - 2x - 2y - 7 = 0
- x² + y² + 2x + 2y + 7 = 0
- x² + y² - 2x - 2y + 7 = 0
Correct answer: x² + y² + 2x + 2y - 7 = 0
Solution
The given circle has centre (2, 3) and radius 5; the radius-3 circle touching it internally at (-1, -1) has the standard textbook equation x² + y² + 2x + 2y - 7 = 0 (centre (-1,-1), radius 3).
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