Question

What is an expression for the radius of the circle? What is the least possible integer value of x for the circle to exist? Explain

Answer 1

The area of a circle is given by the next formula:

$A=\pi *r^2$

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For the given expression of area:

$\pi *r^2=4\pi x^2+12\pi x+9\pi$

Use the equation above to solve r (radius):

1. Divide both sides of the equation into π

$\begin{gathered} \frac{\pi *r^2}{\pi}=\frac{4\pi x^2+12\pi x+9\pi}{\pi} \\ \\ r^2=4x^2+12x+9 \end{gathered}$

2. Factor the expression on the right using the notable product perfect square binomial:

$\begin{gathered} (a+b)\placeholder{⬚}^2=a^2+2ab+b^2 \\ \\ 4x^2+12x+9=(2x)\placeholder{⬚}^2+2(2x)(3)+3^2 \\ 4x^2+12x+9=(2x+3)\placeholder{⬚}^2 \\ \\ \\ r^2=(2x+3)\placeholder{⬚}^2 \end{gathered}$

3. Find square root of both sides of the equation:

$\begin{gathered} \sqrt{r^2}=\sqrt{(2x+3)\placeholder{⬚}^2} \\ \\ r=2x+3 \end{gathered}$

Then, the radius is given by the expression 2x+3.

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To find the least possible integer value of x: As r is the radius of a circle, it cannot be a negative amount or 0, then r needs to be greather than 0:

$\begin{gathered} 2x+3>0 \\ \end{gathered}$

Solve the ineqaulity above:

$\begin{gathered} 2x+3-3>0-3 \\ 2x>-3 \\ \\ \frac{2x}{2}>-\frac{3}{2} \\ \\ x>-\frac{3}{2} \end{gathered}$

x needs to be greater than -3/2 (-1.5).

Then, the least possible integer value of x is -1