Answer:
An equation represents a circle with its center at (2, -3) and that passes through the point (6,2) is:
(x - 2)² + (y + 3)² = 41
Explanation:
The standard equation of a circle having the center at point C(x₀, y₀) and radius r is given by
(x - x₀)² + (y - y₀)² = r²
It means that every point P(x, y) belonging to the circle must satisfy the equation.
In this particular case, we have that the circle passing through P(x, y) = (6, 2) and the center C(x₀, y₀) = (2, -3).
Then substituting these values in the formula:
(x - x₀)² + (y - y₀)² = r²
(6 - 2)² + (2 - (-3))² = r²
switch sides
r² = (6 - 2)² + (2 - (-3))²
r² = (4)² + (2 + 3)²
r² = (4)²+ (5)²
r² = 16 + 25
r² = 41
r = √41, r = -√41
As r can not be negative, so
r = √41
Therefore , the equation of the circle with center at C = (2, -3) and radius r = √41 is:
(x - 2)² + (y + 3)² = (√41 )²
(x - 2)² + (y + 3)² = 41
Therefore, an equation represents a circle with its center at (2, -3) and that passes through the point (6,2) is:
(x - 2)² + (y + 3)² = 41