The area of a circle is given by the next formula:
![A=\pi *r^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/i1im3wds3abxucwfli8zpbankmg3a1nwgl.png)
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For the given expression of area:
![\pi *r^2=4\pi x^2+12\pi x+9\pi](https://img.qammunity.org/2023/formulas/mathematics/high-school/q6flfbdu0x5hrax1skx6m0n0fnkgsag967.png)
Use the equation above to solve r (radius):
1. Divide both sides of the equation into π
![\begin{gathered} (\pi *r^2)/(\pi)=(4\pi x^2+12\pi x+9\pi)/(\pi) \\ \\ r^2=4x^2+12x+9 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2pjwmjcpx58wzrvcsuoac40zbogaikgnja.png)
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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/r99ljobuauuj548iu7vbq6q6t19zuro1f3.png)
3. Find square root of both sides of the equation:
![\begin{gathered} √(r^2)=\sqrt{(2x+3)\placeholder{⬚}^2} \\ \\ r=2x+3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/mnt17hrb61oxiwqefhhyuho7vwwxp7x4co.png)
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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/5q7x4y7o67ad7gcj81jcnx5ipx4sz4z5ci.png)
Solve the ineqaulity above:
![\begin{gathered} 2x+3-3>0-3 \\ 2x>-3 \\ \\ (2x)/(2)>-(3)/(2) \\ \\ x>-(3)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/w0e1p2kgf19u1tipjoyyeysrqegs8q0wpn.png)
x needs to be greater than -3/2 (-1.5).
Then, the least possible integer value of x is -1