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Substitute in each point as $(x,y)$ into the general equation of the circle, obtaining 3 (simultaneous) equations in $a, b$ and $c$ that you must solve.
e.g. since $(-4,1)$ lies on the circle, we know that $(x,y)=(-4,1)$ satisfies $x^2+y^2+ax+by+c=0$.
i.e. $(-4)^2+1^2+a(-4)+b(1)+c=0$.
Tidying this up, we get: $-4a+b+c+17=0 \tag{1}$
Using the other two points, and doing the same think I've just done, we obtain two more equations:
$ 3a+c+9=0\tag{2}$
$ 5a+4b+c+41=0\tag{3}$
Now, solve equations $(1), (2)$ and $(3)$ (simultaneously) for $a, b$ and $c$ and you will have your equation.
You should find that $a=0, b=-8, c=-9$.
Write the equation of the circle as follow: $$ x^2+y^2+ax+by+c=0\Leftrightarrow (x+x_0)^2+(y+y_0)^2=r^2. $$ $x_0=\dfrac{a}{2}$, $y_0=\dfrac{b}{2}$, and $r^2=\dfrac{a^2+b^2-4c}{4}$. Of course, $a^2+b^2-4c\geqslant0$ must holds.
Replace your points $(x, y)$ and you will see if it satisfies the circle's equation or not.