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The line 2x - 3y + 1 = 0 touches a circle S = 0 at the point (1, 1). If the radius of the circle is sqrt(13), find the equation(s) of the circle S.

1. (x-3)² + (y+2)² = 13 or (x+1)² + (y-4)² = 13
2. (x-3)² + (y-2)² = 13 or (x+1)² + (y+4)² = 13
3. (x-2)² + (y+3)² = 13 or (x+2)² + (y-3)² = 13
4. (x-3)² + (y+2)² = 13 only

Correct answer: (x-3)² + (y+2)² = 13 or (x+1)² + (y-4)² = 13

Solution

Stepping a distance sqrt(13) from (1,1) along the unit normal (2,-3)/sqrt(13) gives centres (3,-2) and (-1,4), producing the two circles (x-3)² + (y+2)² = 13 and (x+1)² + (y-4)² = 13.

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