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Question

# Graph the circles whose equations are given. Label each circle's center and intercepts (if any) with their coordinate pairs.$x^2+y^2-4 x-(9 / 4)=0$

Solution

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Step 1
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Write the given equation into the standard equation of circle. The standard equation of a circle is

\begin{align} (x-h)^2 + (y-k)^2 = a^2 \end{align}

Where the center is $(h,k)$ and the radius is $a$.

To write the given equation into the standard equation (1), separate the constant on one side. After that, complete the squares for the $x$ and $y$ variables. Note that to complete a square of the expression $x^2+bx$, get the value equal to $(b/2)^2$ then, add it to the expression $x^2+bx$. From this, the resulting equation will be $(x+b/2)^2$.

\begin{align*} x^2+y^2-4x & = 9/4 \\ (x-4x+4)^2 + (y-0)^2 &= 9/4 + 4\\ (x-2)^2 + (y-0)^2 &= 25/4 \\ (x-2)^2 + (y-0)^2 &= (5/2)^2 \\ \end{align*}

Thus, $a=5/2$ and $(h,k) = (2,0)$.

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