The equation of the circle is D. x² + y² - 2y + 1 = 0.
What is the equation of the circle?
We know the foot of the normal to the circle from the point (4,3) is (2,1).
We also know the equation of a diameter of the circle: 2x - y - 2 = 0.
The foot of the normal lies on the circle and the normal line passes through the foot and the point (4,3).
The slope of the diameter (positive 2) is negative reciprocal of the slope of the normal line.
Slope of normal = -(slope of diameter) = -2
Equation of normal line passing through (2,1) with slope -2: y - 1 = -2(x - 2)
The center of the circle lies on the normal line and is equidistant from the foot and the point (4,3).
The radius of the circle is the distance between the center and the foot of the normal: √((a - 2)² + (b - 1)²) = √5.
Centered at (a,b), with radius √5, the equation of the circle is:
(x - a)² + (y - b)² = 5
Only option D, x² + y² - 2y + 1 = 0, satisfies the equation with the calculated center and radius.
Therefore, the equation of the circle is D. x² + y² - 2y + 1 = 0.