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Two lines 3x - 4y - 7 = 0 and 2x - 3y - 5 = 0 are normals to a circle that passes through the point (2, 0). Find the equation of the circle.

1. x² + y² - 2x + 2y + 2 = 0
2. x² + y² - 2x + 2y = 0
3. x² + y² - 2x = 0
4. x² + y² + 2x + 2y - 8 = 0

Correct answer: x² + y² - 2x + 2y = 0

Solution

Any normal to a circle passes through its centre. So the centre is the intersection of 3x-4y-7=0 and 2x-3y-5=0. After finding the centre, use the given point (2,0) to determine the radius.

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