The vertices of △XYZ are X(4, 5), Y(4, 13), and Z(8, 9). Find an equation of the circle circumscribed about △XYZ.

An equation is ???

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The equation of the circle is (x - 4)² + (y - 9)² = 4²

Explanation:

The center radius form of the equation of the circle is (x - h)² + (y - k)² = r²

Therefore, given the coordinates of the vertices of the inscribed triangle, we have points on the circumference includes;

X(4, 5), Y(4, 13), Z(8, 9)

Hence, we have;

(4 - h)² + (5 - k)² = r² = h²- 8·h - 10·k + k² + 41(1)

(4 - h)² + (13 - k)² = r² = h² - 8·h + k² - 26·k +185(2)

(8 - h)² + (9 - k)² = r² = h² - 16·h -18·k + k² +145(3)

Subtracting equation (1) from (2), we have;

-(16·k -144) = 0

∴ 16 k = 144

k = 9

Subtracting equation (1) from (3), we have;

-(8·h + 8·k - 104) = 0

Which gives;

8·h + 8·k - 104 = 0, where k = 9 gives

8·h + 8×9 - 104 = 0

8·h - 32 = 0

h = 32/8 = 4

Plugging in the values of h and k in equation (3), we have;

(8 - 4)² + (9 - 9)² = r²

4² = r²

∴ r = 4

The equation of the circle is (x - 4)² + (y - 9)² = 4².