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: \[ x^2 - 8x - 14y + y^2 + 61 = 0 \] 

Answer 2

Step 1: Analyze the given equations of the square.

We are given two quadratic equations that represent the boundaries of the square:

• \( x^2 - 8x + 12 = 0 \)
• \( y^2 - 14y + 45 = 0 \)

These equations describe the vertical and horizontal sides of the square. To find the vertices of the square, we solve these equations for \( x \) and \( y \).

Step 2: Solve \( x^2 - 8x + 12 = 0 \).

This is a quadratic equation in \( x \). Factoring:

\[ x^2 - 8x + 12 = (x - 6)(x - 2) = 0 \]

Thus, the solutions are:

\[ x = 6 \quad \text{and} \quad x = 2 \]

Step 3: Solve \( y^2 - 14y + 45 = 0 \).

This is a quadratic equation in \( y \). Factoring:

\[ y^2 - 14y + 45 = (y - 9)(y - 5) = 0 \]

Thus, the solutions are:

\[ y = 9 \quad \text{and} \quad y = 5 \]

Step 4: Determine the vertices of the square.

The square is bounded by the lines \( x = 2 \), \( x = 6 \), \( y = 5 \), and \( y = 9 \). The vertices of the square are:

\[ (2, 5), (2, 9), (6, 5), (6, 9) \]

Step 5: Find the center and radius of the inscribed circle.

The inscribed circle is centered at the midpoint of the square, which is also the midpoint of its diagonals. The midpoint is:

\[ \left( \frac{2 + 6}{2}, \frac{5 + 9}{2} \right) = (4, 7) \]

The radius of the inscribed circle is half the side length of the square. The side length is:

\[ 6 - 2 = 4 \quad \text{(horizontal side)} \quad \text{or} \quad 9 - 5 = 4 \quad \text{(vertical side)}. \]

Thus, the radius is:

\[ r = \frac{4}{2} = 2 \]

Step 6: Write the equation of the circle.

The general equation of a circle is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substitute \( h = 4 \), \( k = 7 \), and \( r = 2 \):

\[ (x - 4)^2 + (y - 7)^2 = 2^2 \]

\[ (x - 4)^2 + (y - 7)^2 = 4 \]

Expand the equation:

\[ (x^2 - 8x + 16) + (y^2 - 14y + 49) = 4 \]

Simplify:

\[ x^2 - 8x + y^2 - 14y + 65 = 4 \]

\[ x^2 - 8x + y^2 - 14y + 61 = 0 \]

Step 7: Match with the given options.

The equation of the circle is:

\[ x^2 - 8x - 14y + 61 = 0 \]

Final Answer:

\( x^2 - 8x - 14y + 61 = 0 \)