Question

The equation of the circle that can be inscribed in the square formed by \(x^2-8x+12=0\) and \(y^2-14y+45=0\) is ?

• A. \(x^2-8x-14y+61=0\)
• B. \(x^2-8x-14y+71=0\)
• C. \(x^2-4x-7y+61=0\)
• D. \(x^2-4x-7y+71=0\)
• E. \(x^2-4x-7y+81=0\)
• F. \[ x^2 - 8x + y^2 - 14y + 61 = 0 \]

Answer 1

Answer: (F)

Step 1: Find the sides of the square from the given equations

The square is formed by the lines:

• \(x^2 - 8x + 12 = 0\) which factors to \((x - 2)(x - 6) = 0\), giving vertical lines at \(x = 2\) and \(x = 6\) (width = \(6 - 2 = 4\) units)
• \(y^2 - 14y + 45 = 0\) which factors to \((y - 5)(y - 9) = 0\), giving horizontal lines at \(y = 5\) and \(y = 9\) (height = \(9 - 5 = 4\) units)

Thus, we have a square with side length 4 units.

Step 2: Determine the inscribed circle's properties

The circle inscribed in the square will:

• Have diameter equal to the side of the square, so diameter = 4 ⇒ radius \(r = 2\)
• Be centered at the intersection of the square's diagonals

Center coordinates: \[ \text{Midpoint of x-values} = \frac{2 + 6}{2} = 4 \] \[ \text{Midpoint of y-values} = \frac{5 + 9}{2} = 7 \] So the center is at \((4, 7)\).

Step 3: Write the equation of the circle

Using the standard form \((x - h)^2 + (y - k)^2 = r^2\) where \((h, k)\) is the center and \(r\) is the radius: \[ (x - 4)^2 + (y - 7)^2 = 2^2 \] Expanding this: \[ x^2 - 8x + 16 + y^2 - 14y + 49 = 4 \] \[ x^2 - 8x + y^2 - 14y + 61 = 0 \] Rearranged form: \[ x^2 - 8x - 14y + y^2 + 61 = 0 \]