Diwali Special

Easier Solution thanks to @Atomsky Jahid

Using the general formula of a circle (x-p)^2 + (y-q)^2 & = r^2 , where $(p,q)$ is the center and $r$ the radius of the circle, then we have:

$\begin{aligned} (x-p)^2 + (y-q)^2 & = r^2 & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ 2(x-p) + 2(y-q)\frac {dy}{dx} & = 0 \\ \frac {dy}{dx} & = - \frac {x-p}{y-q} \\ \frac {ax+3}{2y+f} & = - \frac {x-p}{y-q} & \small \color{#3D99F6} \text{Equating the coefficients on both sides} \\ \implies a & = \boxed{-2} \end{aligned}$

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Previous solution

$\begin{aligned} \frac{dy}{dx} & = \frac{\color{#3D99F6}{a}x+3}{2y+f} \\ (2y+f) \frac{dy}{dx} & = \color{#3D99F6}{a}x+3 \\ \int (2y+f) \ dy & = \int {\color{#3D99F6}{a}x+3} \ dx \\ y^2 + fy & = \frac{\color{#3D99F6}{a}}{2}x^2+3x +C \quad \quad \small \color{#3D99F6}{\text{where } C \text{ is the constant of integration.}} \\ y^2 + fy - \frac{\color{#3D99F6}{a}}{2}x^2 - 3x & = C \quad \quad ...(1) \end{aligned}$

For $(1)$ to be a circle, it is of the form:

$\begin{aligned} (x-p)^2 + (y-q)^2 & = r^2 & \small \color{#3D99F6} \text{where }(p,q) \text{ is the center of the circle and }r \text{, the radius.} \\ {\color{#3D99F6}\left(-\frac a2\right)x^2 - 3x + \frac 94} + {\color{#D61F06} y^2 + fy + \frac {f^2}4} & = \frac 94 + \frac {f^2}4 + C \\ {\color{#3D99F6}\left(x-\dfrac{3}{2}\right)^2} + {\color{#D61F06}\left(y+\dfrac{f}{2}\right)^2} & = \dfrac{9+f^2}{4} + C & \small \color{#3D99F6} \text{Note that the coefficient of }x^2 \text{ must be }1. \\ \implies - \dfrac{\color{#3D99F6}{a}}{2} & = 1 \\ \implies \color{#3D99F6}{a} & = \boxed{-2} \end{aligned}$