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

Another point on the line is (6,-1)...

What is the center of the circle X2 + y2 = 28+ - id222835693865, QueenNerdy889

Ответ:

cheesecake1919

22.01.2022

Mathematics

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 together and anything with a together:

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

Subtract on both sides:

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

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

We need to complete the square for 's. The '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 '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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What is the center of the circle X2 + y2 = 28+ 3x?

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