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Tamiku [17]
3 years ago
13

Determine the equation of a circle having a diameter with endpoints at (12,-14) and(2,4)

Mathematics
1 answer:
snow_lady [41]3 years ago
3 0
Center of a circle is a midpoint of line connecting two endpoints so center is:
(12,-14) (2,4) we divide those two points by two 
C(a,b)- center
C(a,b)=((12+2)/2),(-14+4)/2)=(7,-5)
Radius is the distance from the center to any point on the circle:
we take point (2,4)
r=\sqrt{ (7-2)^{2} + (-5-4)^{2} } = \sqrt{25+81} = \sqrt{106}

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Answer:

y = (1/2)x² -2x

Step-by-step explanation:

recall that the general equation of a quadratic function is:

y = Ax² + Bx + C

given that 0 and 4 are roots, that means when x = 0 and x = 4, then  y = 0

From this we can get 2 points on the curve, namely,  (0,0) and (4,0)

substituting these points one at a time into the equation above,

for the first point (0,0),

0 = A (0)² + b(0) + C

C = 0

hence the equation becomes:    y = Ax² + Bx

for the 2nd known point (4,0)

0 = A(4)² + B(4)

0 = 16A + 4B   (divide both sides by 4)

0 = 4A + B

B = -4A    ------------(eq 1)

we are given a 3rd point, vertex at (2,-2)

for (2,-2)

-2 = A(2²) + B(2)

-2 = 4A + 2B   (divide both sides by 2)

-1 = 2A + B   (subtract 2A from both sides)

B = -1 -2A --------(eq 2)

solving the system of equations using the method of your choice in eq1 and eq 2 gives:

A = 1/2 and B = -2

hence the equation is

y = (1/2)x² -2x

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Answer:

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Step-by-step explanation:

7 0
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Read 2 more answers
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3x+2y=7\ \ \ |subract\ 2y\ from\ both\ sides\\\\3x=7-2y\ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{x=\dfrac{7-2y}{3}}
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