If the ends of a diameter are A(x1, y1) and B(x2, y2), the equation of the circle is?
A
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
B
(x - h)² + (y - k)² = r²
Analysis & Theory
When ends of a diameter are given, the circle’s equation is (x - x1)(x - x2) + (y - y1)(y - y2) = 0.
The center of the circle with ends of diameter A(2, 4) and B(6, -2) is?
Analysis & Theory
Center = midpoint = ((2+6)/2, (4+(-2))/2) = (4, 1).
The radius of the circle with diameter joining (1, 2) and (5, 6) is?
Analysis & Theory
Length of diameter = √[(5-1)² + (6-2)²] = √(16+16) = √32 = 4√2. Radius = half = 2√2.
Equation of circle with ends of diameter (0, 0) and (4, 0) is?
C
(x - 4)(x - 0) + y² = 0
Analysis & Theory
Center = (2, 0), radius = 2. Equation: (x - 2)² + y² = 4.
Equation of circle with ends of diameter (1, 2) and (3, 4) is?
A
(x - 1)(x - 3) + (y - 2)(y - 4) = 0
B
(x - 2)² + (y - 3)² = 2
Analysis & Theory
Equation can be written as (x - 1)(x - 3) + (y - 2)(y - 4) = 0 or in center-radius form with center (2,3) and radius √2.
If the ends of a diameter are (–2, 0) and (2, 0), then the circle’s equation is?
Analysis & Theory
Diameter = 4 ⇒ radius = 2, center at (0,0). Equation: x² + y² = 4.
For ends of diameter (–a, 0) and (a, 0), the circle equation is?
Analysis & Theory
Center = (0,0), radius = a. Equation: x² + y² = a².
If the ends of a diameter are A(0, –3) and B(0, 3), the circle equation is?
Analysis & Theory
Diameter = 6 ⇒ radius = 3 ⇒ equation: x² + y² = 9.
Equation of circle with ends of diameter (–1, –1) and (3, 5) is?
A
(x + 1)(x - 3) + (y + 1)(y - 5) = 0
B
(x - 1)² + (y - 2)² = 10
Analysis & Theory
Equation can be written in product form (x+1)(x-3) + (y+1)(y-5) = 0 or standard form with center (1,2), radius √10.
If the ends of a diameter are (x1, y1) and (x2, y2), the center is?
C
((x1 + x2)/2, (y1 + y2)/2)
Analysis & Theory
The center of the circle is the midpoint of the diameter = ((x1 + x2)/2, (y1 + y2)/2).