Get the equation of a circle when given 3 points Get the equation of a circle through the points $(1,1), (2,4), (5,3) $.
I can solve this by simply drawing it, but is there a way of solving it (easily) without having to draw?

Based on what I've learnt from this page by Stephen R. Schmitt (thanks to @Dan_Uznanski for pointing the concept out), there's a faster way to do it using matrices:
$$\text{let }〈x_1, y_1〉, 〈x_2, y_2〉, 〈x_3, y_3〉\text{ be your 3 points, and let }〈x_0, y_0〉\text{ represent the center of the circle.}\\\
\\
\text{let }A =
\left[\begin{array}{cccc}
    x^2+y^2 & x & y & 1\\
    x_1^2+y_1^2 & x_1 & y_1 & 1\\
    x_2^2+y_2^2 & x_2 & y_2 & 1\\
    x_3^2+y_3^2 & x_3 & y_3 & 1\\
\end{array}\right] =
\left[\begin{array}{cccc}
    x^2+y^2 & x & y & 1\\
    1^2+1^2 & 1 & 1 & 1\\
    2^2+4^2 & 2 & 4 & 1\\
    5^2+3^2 & 5 & 3 & 1\\
\end{array}\right] =
\left[\begin{array}{cccc}
    x^2+y^2 & x & y & 1\\
    2 & 1 & 1 & 1\\
    20 & 2 & 4 & 1\\
    34 & 5 & 3 & 1\\
\end{array}\right]\\\ \\\
\\\
\text{Note: }M_{11} = 0 ⟹ 〈x_0, y_0〉\text{ is undefined}∧\text{the points lie on a line rather than a circle.}\\\
\\\
x_0 =
\frac{1}{2} ⋅ \frac{M_{12}}{M_{11}}
= \frac{1}{2} ⋅ 
\left|\begin{array}{cccc}
    2 & 1 & 1\\
    20 & 4 & 1\\
    34 & 3 & 1\\
\end{array}\right|
\Bigg{/}
\left|\begin{array}{cccc}
    1 & 1 & 1\\
    2 & 4 & 1\\
    5 & 3 & 1\\
\end{array}\right|
=
\frac{1}{2} ⋅ \frac{-60}{-10} = 3
$$
$$
y_0 =
-\frac{1}{2} ⋅ \frac{M_{13}}{M_{11}}
= -\frac{1}{2} ⋅
\left|\begin{array}{cccc}
    2 & 1 & 1\\
    20 & 2 & 1\\
    34 & 5 & 1\\
\end{array}\right|
\Bigg{/}
\left|\begin{array}{cccc}
    1 & 1 & 1\\
    2 & 4 & 1\\
    5 & 3 & 1\\
\end{array}\right|
=
-\frac{1}{2} ⋅ \frac{40}{-10} = 2
$$
The radius, $r$, can then be calculated with Pythogoras' theorem, or using matrices again:
$$r^2 = x_0^2 + y_0^2 + \frac{M_{14}}{M_{11}}$$

*

*$M_{12}(A)$ is a minor of A — the determinant of $A$ without row $1$ or column $2$.

*Lines on either side of a matrix indicate the determinant (e.g. $|A|$) (sometimes written $\det{(A)}$)


If you have the time, it's really worth learning about matrices — they make a lot of things much faster, and are just generally awesome! KhanAcademy has a pretty easy introduction to matrices and linear algebra.
