Here is the link to the playlist of videos that explain all of the math questions in the 2017 October SAT Question and Answer Service test. The playlists are separated by sections. Section 3 of the math section does not allow use of calculator, whereas the calculator is permitted in Section 4 of the math section.

## 2017 PSAT Wednesday Test: Video explanations

Here is the link to the playlist of videos that explain all of the math questions in the 2017 PSAT Wednesday test that was conducted on the 11th of October: **2017 PSAT Wednesday Test Video Explanations**.

## SAT math practice question: Dividing algebraic expressions

Try the following SAT math question that deals with algebraically dividing expressions.

Which of the following is equivalent to $\dfrac{x^2-2x-3}{x+2}$ ?

- $\quad x \, -\, \dfrac{3}{x+2}$
- $\quad x \,- \, \dfrac{7}{x+2}$
- $\quad x \, – 4 \, + \dfrac{5}{x+2}$
- $\quad x + 4 \; – \; \dfrac{11}{x+2}$

## 2017 May SAT Test: Video explanations

Here is the link to the playlist of videos that explain all of the math questions in the 2017 May SAT test.

## Equation of a circle in the expanded form

**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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