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blagie [28]
3 years ago
15

Find the center and radius of the circle with the equation x^2 + y^2+8x-4y=-11

Mathematics
2 answers:
S_A_V [24]3 years ago
8 0
x^2 + y^2+8x-4y=-11 \\ \\ x^2+8x + y^2 -4y =-11\\ \\ (x^2+8x) + ( y^2 -4y) =-11\\ \\(x^2+8x+16) + ( y^2 -4y+4)-16-4 =-11 \\ \\(x+4)^2+(y-2)^2-20=-11

(x+4)^2+(y-2)^2 =-11+20\\ \\(x+4)^2+(y-2)^2 =9 \\ \\x+4=0 \ \ and \ \ y-2 =0 \\ \\ x=-4 \ \ and \ \ y=2 \\ \\ So \ the \ center \ of \ the \ circle \ is (-4, 2)

The \ radius \ of \ the \ circle \ is \ the \ square \ root \ of \ the \ right \\ \\ side \ of \ the \ equation \ for \ this \ circle. \\ \\ So\ the \ radius \ of \ this \ circle \ is \ \sqrt{9} \ so \ the \ radius \ of \ the \ circle \ is \ 3


hoa [83]3 years ago
5 0
X^2+8x+16+y^2-4y+4=-11+16+4
[x+4]^2+[y-2]^2=9
radius is 3 and center is [-4,2]
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I need help with number 3
yan [13]
First check whether the point (-6,8) is the solution to any of the equations. To check, just plug in the x and y values of the points into the equation and see if they give you a true statement. 
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A. It is the only solution to the set. 
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3 years ago
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The number in each column are related. The goal is to determine how they are related, determine which numbers belongs in the bla
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Answer:

If you cube the numbers in the left column, you get the numbers in the right column! We can figure this out by understanding that 1^3 = 1, 3^3 = 27, and 6^3 = 216. Then it all falls into place!

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3 years ago
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