What is the area of a circle whose circumference passes through the points of 6 3 and -5 2 and 7 2 on the Cartesian plane showing work?

Answer 1

A circle with centre (X, Y) and radius r has formula:

(x - X)² + (y - Y)² = r²

As the points (6, 3), (-5, 2) and (7, 2) are on the circle:

1. (6 - X)² + (3 - Y)² = r²
2. (-5 - X)² + (2 - Y)² = r²
3. (7 - X)² + (2 - Y)² = r²

Subtracting (2) from (3) gives:

(7 - X)² - (-5 - X)² + (2 - Y)² - (2 - Y)² = r² - r²

→ 49 - 14X + X² - (25 +10X +X²) = 0

→ 49 - 25 -14X -10X +X² - X² = 0

→ 24 - 24X = 0

→ 24X = 24

→ X = 1

Substituting this into (1) and (2) gives:

1. (6 - 1)²+ (3 - Y)² = r² → 5² + 9 - 6Y + Y² = r² → 34 - 6Y + Y² = r²
2. (-5 - 1)² + (2 - Y)² = r² → (-6)² + 4 - 4Y + Y² = r² → 40 - 4Y +Y² = r²

Subtracting (1) from (2) now gives:

40 - 34 - 4Y - (-6Y) +Y² - Y² = r² - r²

→ 6 + 2Y = 0

← 2Y = -6

→ Y = -3

→ The equation of the circle is (x - 1)² + (y - -3)² = r² → (x - 1)² + (y + 3)² = r²

Using one of the points (6, 3), say, and substituting for x and y gives:

(6 - 1)² + (3 + 3)² = r²

→ 5² + 6² = r²

→ 25 + 36 = r²

→ r² = 61

The area of a circle is given by area = π × radius²

→ the area of the circle through those three points is π × 61 = 61π units² ≈ 191.64 units²