Multiple choice

A pair of straight lines $x^{2} - 8x + 12 = 0$ and $y^{2} - 14y + 45 = 0$ are forming a square. What is the centre of circle inscribed in the square.

  1. $(3, 2)$
  2. $(7, 4)$
  3. $(4, 7)$
  4. $(0, 1)$
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C 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. These lines form a square with vertices (2,5), (6,5), (6,9), and (2,9). The center of the inscribed circle is the center of the square, which is the midpoint of the diagonals: ((2+6)/2, (5+9)/2) = (4, 7).

AI explanation

Factoring the equations gives the vertical lines x = 6 and x = 2, as well as the horizontal lines y = 9 and y = 5. The centre of the inscribed circle of this square is the intersection of its diagonals, found by taking the midpoint of the x-values and the midpoint of the y-values: ((6+2)/2, (9+5)/2). This calculates to (4, 7).

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