4 What is the general form of the equation (x-3)²+(y+2)² =25?
A. x²+y²-6x+4y-12=0
B.x2-y2-6x+4y-12=0
C. x2+y2-6x-4y-12=0
D. 22-32-6x-17-12-0
Solution:
(x-3)²+(y+2)² =25
x² - 6x + 3² + y² + 4y + 2² = 25 - 3² - 2²
x² + y² - 6x + 4y = 25 - 9 - 4
x² + y² - 6x + 4y = 12
x² + y² - 6x + 4y - 12 = 0
5. A radius of a circle has endpoints (-4,3) and (1, - 2). What is the equation that defines the circle if its center is at the second quadrant?
A.(X-1)2+(y+2)2=50
B. (x+1)+(y-2)2=50
C.(x+4)²+(y-3)²=50
D.(x-4)2-(y+3y=50
Solution:
(x₁, y₁), (x₂, y₂)
(-4, 3), (1, -2)
d² = (x₂ - x₁)² + (y₂ - y₁)²
d² = (1 - (-4))² + (-2 - 3)²
d² = (5²) + (-5)²
d² = 25 + 25
d² = 50
d = √50
the problem states that the center is at the second quadrant.
therefore, (-4, 3) is the center
(x - h)² + (y - k)² = r²
(x + 4)² + (y - 3)² = (√50)²
(x + 4)² + (y - 3)² = 50
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