Multiple choice

The centre of the circle inscribed in a square formed by the lines $x^2 - 8x + 12 = 0$ and $ y^2 - 14y + 45 = 0$ is

  1. $(4, 7)$
  2. $(7, 4)$
  3. $(9, 4)$
  4. $(4, 9)$
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A Correct answer
Explanation

The lines x^2 - 8x + 12 = 0 factor to (x-2)(x-6)=0, so x=2 and x=6. The lines y^2 - 14y + 45 = 0 factor to (y-5)(y-9)=0, so y=5 and y=9. This forms a square with vertices (2,5), (6,5), (6,9), (2,9). The center is the midpoint of the diagonals, which is ((2+6)/2, (5+9)/2) = (4, 7).

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