<s>[INST] How to get the equation of a circle with the given information You have a center of a circle $M(-2,3)$ going through a point $P(1,7)$.
What is the equation of the circle?
I thought you could solve it by $R^2 = 3^2 + 4^2$, but that would just give a radius of 5 because there aren't any $x$ or $y$'s. Then I thought, should you then add a point $P(x,y)$? I'm not too sure, can someone give me hints?
 [/INST] The equation of a circle, centre $(p,q)$ and radius $r$, is 
$$ (x-p)^2 + (y-q)^2 = r^2 \, . $$
Here $p$, $q$ and $r$ will be fixed numbers and the circle is the set of points $(x,y)$ for which the equation holds, for example: $(x-1)^2 + (y+2)^2 = 4$ is the equation of the circle with centre $(1,-2)$ and radius $2$.
In your case, the centre is $(-2,3)$. There are lots of circles centred at $(-2,3)$. They have equations $(x+2)^2 + (y-3)^2 = r^2$ where $r > 0$ is the radius of the circle. You also know that the circle passes through the point $(1,7)$. We can use this to find the radius. The circle $(x+2)^2 + (y-3)^2 = r^2$ passes through the point $(1,7)$ if and only if $(x,y) = (1,7)$ is a solution to the equation $(x+2)^2 + (y-3)^2 = r^2$. Let's put $(x,y) = (1,7)$ into the equation:
$$(1+2)^2 + (7-3)^2 = r^2 \iff 3^2 + 4^2 = r^2 \iff 25 = r^2 \, . $$
Since $r>0$, we have $r = 5$ (instead of $r = -5$). It follows that your circle has the equation
$$(x+2)^2 + (y-3)^2 = 25 \, . $$
It is the circle centred at $(-2,3)$ with radius $5$.
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