**Key Concept**: If you are given the equation of a circle in the expanded form $$ x^2 – 2hx + y^2 – 2ky = f $$ then the center of the circle is $(h, k)$ and the radius is given by $r = \sqrt{f+h^2+k^2}$.

The above result saves the effort of completing the square and turning the above equation into the general equation of a circle. We will apply this concept to solve a difficult SAT math problem.

**Practice Problem**: The equation of a circle in the $xy$-plane is shown below. What is the radius of the circle?

$$ 2x^2 – 2x + 2y^2 + 10y = 5 $$

- $\sqrt{5}$
- $3$
- $4$
- $5$

**Solution**: First divide the equation by 2

$$ 2x^2 – 2x + 2y^2 + 10y = 5 $$

and rewrite in a form where the coefficients of $x^2$ and $y^2$ terms are equal to 1. We obtain

$$x^2 – x + y^2 + 5y = \dfrac{5}{2}$$

If we compare this to the general expanded form for the equation of a circle

$$ x^2 – 2hx + y^2 – 2ky = f$$

we obtain $-2h=-1$ and $-2k=5$, therefore $h=\dfrac{1}{2}$ and $k=-\dfrac{5}{2}$, and the center of the circle is

$\left(\dfrac{1}{2}, -\dfrac{5}{2}\right)$.

Also, $f=\dfrac{5}{2}$, and we know that $r = \sqrt{f+h^2+k^2}$, where $f=\dfrac{5}{2}$, $h=\dfrac{1}{2}$, and $k=-\dfrac{5}{2}$. On substitution, we obtain $$r = \sqrt{\dfrac{5}{2} + \left(\dfrac{1}{2}\right)^2 + \left(-\dfrac{5}{2}\right)^2} = \sqrt{\dfrac{5}{2} + \dfrac{1}{4} + \dfrac{25}{4}} = \sqrt{\dfrac{10+1+25}{4}} = \sqrt{\dfrac{36}{4}}=\sqrt{9}=3$$

**Explanation**: The result stated above can be obtained by starting with the general equation of a circle with center at $(h, k)$ and a radius of $r$ units:

$$

\begin{align}

(x-h)^2 + (y-k)^2 & = r^2 && \text{Square the algebraic terms}\\

x^2 – 2hx + h^2 + y^2 -2ky + k^2 & = r^2 && \text{Move the constant terms to the right} \\

x^2 – 2hx + y^2 -2ky & = r^2 – h^2 – k^2 && \\

\end{align}

$$

Compare the above equation to the general expanded form

$$ x^2 – 2hx + y^2 – 2ky = f $$

and we find that the coordinates of the center $(h, k)$ appear as part of the coefficients of the $x$ and $y$ terms. Further, $f = r^2 – h^2 – k^2$, which on rearrangment gives $r = \sqrt{f+h^2+k^2}$.

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