Question

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

Answer 1

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})^2

So 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 $\sqrt{\frac{121}{4}}=\frac{11}{2}$.