Which equation represents a circle with its center at (2, -3) and that passes through the point (6,2)?

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

Step-by-step 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  

the answer is d

step-by-step explanation:

21

step-by-step explanation: