Determine where a point lies in relation to a circle, is my answer right? The circle $C$ has equation
$$x^2-8x+y^2+6y=24$$
I completed the square:
$$(x-4)^2+(y+3)^2=49$$
Therefore the radius is 7. I want to know if the point $(9,2)$ is inside the circle. Substituting into the distance formula:
$$\sqrt{(9-4)^2 + (2+3)^2}=\sqrt{50}=7.07\dots$$
So the point lies outside the circle, or have I gone wrong?
Perfectly fine.

However, do note you didn't actually need to compute the square root. Since for all positive real numbers $x&gt;y$ if and only if $x^2&gt;y^2$, it's enough to observe that $50&gt;49$, and that therefore $\sqrt{50}&gt;\sqrt{49}$.
Too complicated.

Simply re-write the given equation in the form  $f(x, y) = 0$

That is,  $x^2-8x+y^2+6y-24 = 0$

$d = \sqrt {f(x_o, y_o)}$ is the "length" of tangent from $P(x_o, y_o)$ to $C:f(x, y) = 0$.

Revised:- 

If  ${f(x_o, y_o)} \gt 0$, that tangent is a real one and P is outside of $C$.

If ${f(x_o, y_o)} \lt 0$, that tangent is an imaginary one and P is inside $C$.

If ${f(x_o, y_o)} = 0$; guess what.......