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enot [183]
3 years ago
5

The circle below is centered at the point (1, 2) and has a radius of length 2. What is its equation?

Mathematics
2 answers:
Alina [70]3 years ago
7 0
The\ equation\ of\ the\ circle:(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-the\ center;\ r-the\ radius\\-------------------------\\\\(1;\ 2)\to a=1\ and\ b=2;\ r=2\\\\subtitute:\\\\(x-1)^2+(y-2)^2=2^2\\\\\boxed{(x-1)^2+(y-2)^2=4}
statuscvo [17]3 years ago
5 0

Quite frankly, I don't see any circle below.

But the equation of the circle that you've described is . . .

         <em>(x - 1)² + (y - 2)²  =  4</em> .


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Answer:

\bf x_1 = 2\\x_2 = 4\\x_3=6\\x_4=8

Step-by-step explanation:

The given interval : x = 0 to x = 10

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\textbf{Therefore, }\bf x_1 = 0 + 2 = 2\\x_2 = 2 +2 =4\\x_3=4+2=6\\x_4=6+2=8

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