107. Rectangle ABCD is drawn in the Cartesian plane with sides parallel to the x- and y-axes and one diagonal through the points B(12, 12) and D(8, 6). Find the equation of the circle with diameter AC.

A. [tex]\((x-8)^2 + (y-6)^2 = 25\)[/tex]

B. [tex]\((x-12)^2 + (y-2)^2 = 5\)[/tex]

C. [tex]\((x-10)^2 + (y-9)^2 = 13\)[/tex]

D. [tex]\((x-9)^2 + (y-10)^2 = \sqrt{13}\)[/tex]



Answer :

To find the equation of the circle with diameter AC, given the points B(12, 12) and D(8, 6) which form a diagonal of rectangle ABCD, follow these steps:

1. Determine the center of the circle:

Since B and D are points on the diagonal, the center of the circle (which is also the midpoint of the rectangle's diagonal) can be found using the midpoint formula:
[tex]\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \][/tex]
Substituting B(12, 12) and D(8, 6) into the formula:
[tex]\[ M = \left(\frac{12 + 8}{2}, \frac{12 + 6}{2}\right) = (10, 9) \][/tex]

2. Calculate the radius of the circle:

The diameter of the circle is the length of the diagonal BD. The radius is half of this length. To find the length of the diagonal BD, use the distance formula:
[tex]\[ BD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting B(12, 12) and D(8, 6):
[tex]\[ BD = \sqrt{(12 - 8)^2 + (12 - 6)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \][/tex]
The radius [tex]\( r \)[/tex] is half of BD:
[tex]\[ r = \frac{\sqrt{52}}{2} = \sqrt{13} \][/tex]

3. Form the equation of the circle:

The general form of the equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\( r \)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
Here, the center is (10, 9) and the radius is [tex]\( \sqrt{13} \)[/tex]:
[tex]\[ (x - 10)^2 + (y - 9)^2 = (\sqrt{13})^2 \][/tex]
Simplify the equation:
[tex]\[ (x - 10)^2 + (y - 9)^2 = 13 \][/tex]

Thus, the equation of the circle with diameter AC is:
[tex]\[ (x - 10)^2 + (y - 9)^2 = 13 \][/tex]

The correct answer is C) (x - 10)^2 + (y - 9)^2 = 13.