## Equation of a circle in the expanded form

**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 $r = \sqrt{f+h^2+k^2}$.

## Free eight official SAT tests from Collegeboard

**SAT Practice Test 8**(same as January 2017 SAT Test)

**SAT Practice Test 7**(same as October 2016 SAT Test)

**SAT Practice Test 6**(same as April 2016 School day SAT Test)

**SAT Practice Test 5**(same as May 2016 SAT Test)

**SAT Practice Test 4**

**SAT Practice Test 3**

**SAT Practice Test 2**

**SAT Practice Test 1**

## 2017 January SAT Test/Practice Test 8: Video explanations

## Quadratic Equations: All you need to know for the SAT Math

**Quadratic Equations**

A quadratic equation has the form \(ax^2 + bx + c=0\), where \(a\), \(b\), and \(c\) are real numbers, and \(a \neq 0\). The $ax^2$ term is called the quadratic term, the $bx$ is the linear term, and $c$ is called the constant term. Consider the quadratic equation \(x^2 - x -6=0\), if we substitute \(x=3\) in this equation we find that the equation holds true because $3^2-3-6=9-9=0$. Similarly if we substitute $x=-2$ in this equation we again find that $ (-2)^2 - (-2) - 6 = 4+2-6=0$. These two values of $x$ are the only two numbers that satisfy this quadratic equation, and are called the roots of the quadratic equation. Sometimes the roots are also referred to as zeros of the quadratic equation, in other words the value of the quadratic expression is zero at these values.

**Factoring Quadratics: Trial and Error**

**Factoring Quadratics: General Case**

- Multiply the coefficient of \(x^2\), $2$ in this case, and the constant term of $-33$, which gives a value of \(-66\).
- Use trial and error to find two numbers that multiply to give \(-66\) but add up to the coefficient of \(x\), which is \(-5\).
- The two numbers are \(-11\) and \(6\).
- Rewrite the middle term as the sum of these two numbers \(-5x = -11x+6x\).
$$ 2x^2-5x-33=2x^2-11x + 6x-33=0 $$
Factor the largest common term from the first two terms, and also from the last two terms. $$ x(2x-11) + 3(2x-11)=0$$ $$(2x-11)(x + 3)=0 \quad \text{Factor \((2x-11)\)} $$ - The roots are then obtained by solving \(2x-11=0\) and \(x+3=0\), which gives \(\dfrac{11}{2}\) and \(-3\) as the two roots of the quadratic equation \(2x^2-5x-33=0\).

**Completing a Square**

In general if we are given a quadratic of the form $x^2+bx$, we add $\left(\dfrac{b}{2}\right)^2$ to the quadratic to obtain a perfect square. $$ x^2 + bx + \left(\dfrac{b}{2}\right)^2 = x^2+ bx + \dfrac{b^2}{4} = \left(x+ \dfrac{b}{2}\right)^2$$

**Quadratic Expressions: Maximum and Minimum**

**Solved Example:**
$$ h(t) = -16t^2+24t+18$$
The equation above gives the height above ground $h$, in feet, of a ball $t$ seconds after it is launched
vertically upward from a platform that is at a height of 18 feet. What is the greatest height the ball achieves?

- $21$
- $24$
- $27$
- $30$

**Explanation:**
The easiest way to solve this problem is by recognizing that the maximum value of the expression occurs when
$t = -\dfrac{b}{2a}= -\dfrac{24}{2(-16)} = \dfrac{3}{4}$, and we can find the
maximum height by substituting $t=\dfrac{3}{4}$ in the expression $-16t^2+24t+18= -16\left(\dfrac{3}{4}\right)^2 + 24\left(\dfrac{3}{4}\right) + 18
= -9 + 18 + 18 = 27$.
We can also do this problem by completing the square, we first factor out the term $-16$ from the first two terms of the quadratic
expression to obtain:
$$
\begin{align}
h(t) &= -16t^2+24t+18 && \text{Factor out $-16$}\\
&= -16\left[t^2 - \dfrac{3}{2}\right] + 18 && \text{Complete the square} \\
&=-16\left[t^2 - \dfrac{3}{2}+ \left(\dfrac{3}{4}\right)^2 - \left(\dfrac{3}{4}\right)^2\right] + 18 && \text{} \\
&= -16\left[t^2 - \dfrac{3}{2}+ \left(\dfrac{3}{4}\right)^2\right] +16\left(\dfrac{3}{4}\right)^2+ 18 && \text{}\\
& = -16\left(t-\dfrac{3}{4}\right)^2 + 27 && \\
\end{align}
$$
Note that the expression $\left(t-\dfrac{3}{4}\right)^2 \geq 0$ because it is a square of a number. This means that the term $-16\left(t-\dfrac{3}{4}\right)^2$
will always be less than or equal to zero. The maximum height will have a value of $27$ and this will happen when $t=\dfrac{3}{4}$.

**Quadratic Formula**

**Example**: \(2x^2-5x-33=0\), we have \(a=2\), $b=-5$, and \(c=-33\), replacing these values in the above quadratic formula yields $$ x = \dfrac{-(-5) \pm \sqrt{(-5)^2-4(2)(-33)}}{2(2)}=\dfrac{5 \pm \sqrt{289}}{4} = \dfrac{5 \pm 17}{4} $$ and the two solutions are $x=\dfrac{11}{2}$ and $x=-3$.

**Solved Example:**
$$ \dfrac{x^2}{2} = mx + \dfrac{n}{2} $$
In the quadratic equation above, $m$ and $n$ are constants. What are the solutions for $x$ ?

- $\quad m \pm \sqrt{m^2-n}$
- $\quad m \pm \sqrt{m^2+n}$
- $\quad -m \pm \sqrt{m^2-n}$
- $\quad -m \pm \sqrt{m^2+n}$

**Explanation:**
$$
\begin{align}
\dfrac{x^2}{2} &= mx + \dfrac{n}{2} && \text{Multiply both sides by 2}\\
x^2 &= 2mx +n && \text{Rewrite in standard form} \\
x^2 - 2mx - n &=0 && \text{Note $a=1$, $b=-2m$, $c=-n$} \\
x &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} && \text{Apply quadratic formula}\\
& = \dfrac{-(-2m) \pm \sqrt{(-2m)^2-4(1)(-n)}}{2(1)} && \\
& = \dfrac{2m \pm \sqrt{4m^2+4n}}{2} &&\\
& = \dfrac{2m \pm 2\sqrt{m^2+n}}{2} && \text{Divide throughout by 2}\\
& = m \pm \sqrt{m^2+n} &&\\
\end{align}
$$

**Sum and Product of Roots**

**Solved Example:**
$$ 2x^2=3x+2 $$
What is the sum of all the solutions to the quadratic equation above?

- $\quad -\dfrac{5}{2}$
- $\quad -\dfrac{3}{2}$
- $\quad \quad \dfrac{5}{2}$
- $\quad \quad \dfrac{5}{2}$

**Explanation:** Rewrite the equation in the standard quadratic form $ax^2+bx+c=0$ as $2x^2-3x-2=0$. We have $a=2$, $b=-3$, and $c=-2$.
The sum of the roots is equal to $-\dfrac{b}{a}=-\dfrac{(-3)}{2} = \dfrac{3}{2}$. There is no need to waste time in first finding the two roots, which are
$-\dfrac{1}{2}$ and $2$, and adding them to find their sum of $\dfrac{3}{2}$.

**Nature of Roots**

**Example:**
If $b$ is a constant such that the equation $2x^2=bx-3$ has two distinct real solutions, which of the following could be the
value of $b$ ?

- $\quad -5$
- $\quad -4$
- $\quad -3$
- $\quad 4$

**Explanation:** First write the quadratic equation in the standard form $2x^2-bx+3=0$. For the roots to
be real and distinct $b^2-4ac>0$, which is equivalent to $(-b)^2 - 4(2)(3)>0$ or $b^2>24$. The only value in the answer
choices that satisfies this condition is $-5$.