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jonny [76]
3 years ago
15

What is the center of the circle X2 + y2 = 28+ 3x?

Mathematics
1 answer:
notka56 [123]3 years ago
5 0

Answer:

The center is (\frac{3}{2},0).

The problem:

What is the center of the circle X2 + y2 = 28+ 3x?

Step-by-step explanation:

We need to put anything with an x together and anything with a y together:

x^2+y^2=28+3x

Subtract 3x on both sides:

(x^2+y^2)-3x=(28+3x)-3x

x^2-3x+y^2=28

We need to complete the square for x's. The y's are already done.

Keep in mind: x^2+bx+(\frac{b}{2})^2=(x+\frac{b}{2})^2So we are going to add [tex](\frac{-3}{2})^2 on both sides so we can use this identity for the left side with the x's:

x^2-3x+(\frac{-3}{2})^2+y^2=28+(\frac{3}{2})^2

(x+\frac{-3}{2})^2+y^2=28+\frac{9}{4}

(x-\frac{3}{2})^2+y^2=\frac{28(4)+9}{4}

(x-\frac{3}{2})^2+(y-0)^2=\frac{121}{4}

So the center is (\frac{3}{2},0).

The radius is \sqrt{\frac{121}{4}}=\frac{11}{2}.

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