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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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olchik [2.2K]

Answer:

Exact Form:

x = 0, 1/4

Decimal Form:

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Step-by-step explanation:

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Factor 3x^2 out of -3x^2 <em>(4x) + 3x^2 (-1) </em>:

<em> 3x^2 (4x-1) = 0</em>

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<u>Step 2: Divide each term by 3 and simpify</u>

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divide each term in 3x^2 (4x-1) = 0 by 3.

3x^2 (4x-1) / 3 = 0 / 3

<em>simplify 3x^2 (4x-1) / 3.</em>

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<em>Cancel the common factors.</em>

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divide x^2 (4x-1) by 1.

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multiply x^2 by x^2 by adding the exponents.

Move x

4 (x · x^2) -1 · x^2= 0 / 3

Multiply x by x^2

Rase x to the power of 1.

4 (x^1 · x^2) -1 · x^2= 0 / 3

Use the power rule a^m a^n = a^m+n to combine exponents

4x^1+2 -1 · x^2= 0 / 3

Add 1 and 2.

4x^3 -1 · x^2= 0 / 3

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4x^3 -x^2= 0 / 3

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