What is the area of a circle whose circumference passes through the points of 6 2 and 8 -2 and 0 2 on the Cartesian plane showing work?

Answer 1

Points: (6, 2) (8, -2) (0, 2) The Cartesian equation works out as (x-3)^2 +(y+2)^2 = 25 Therefore the circle's centre is at (3, -2) and its radius is 5 Check: distance from (3, -2) to any of the given points is 5 which is the radius Area of the circle: pi*5 squared = 25*pi or about 78.539 square units

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The equation of a circle with centre (X, Y) and radius r is given by:

(x - X)² + (y - Y)² = r²

The three points (6, 2), (8, -2) and (0, 2) are on the circle therefore:

1. (6 - X)² + (2 - Y)² = r²
2. (8 - X)² + (-2 - Y)² = r²
3. (0 - X)² + (2 - Y)² = r²

Expanding each of these gives:

1. 36 - 12X + X² + 4 - 4Y + Y² = r²
2. 64 - 16X + X² + 4 + 4Y + Y² = r²
3. X² + 4 - 4Y + Y² = r²

Subtracting equation 3 from 1 gives:

36 -12X = 0

→ 12 X = 36

→ X = 3

Substituting for X = 3 in the equations gives:

1. 36 - 36 + 9 + 4 - 4Y + Y² = r²
2. 64 - 48 + 9 + 4 + 4Y + Y² = r²
3. 9 + 4 - 4Y + Y² = r²

Subtracting equation 3 from 2 gives:

16 + 8Y = 0

→ 8Y = -16

→ Y = -2

Thus the circle has equation (x - 3)² + (y + 2)² = r²

Using point (0, 2) gives:

(0 - 3)² + (2 + 2)² = r²

→ r² = 9 + 16 = 25

The area of a circle is given by

area = πr² = 25π ≈ 78.54 sq units.