How to calculate the centre of a circle given two points and the equation of a line going through the centre? A(3,5) and b(9,-3) lie on a circle. show that the centre of the circle lies on the line 4y-3x+14 = 0
Is it possible to work out the centre of the circle algebraically since the only method I can think of is drawing the points and the line on a set of axis and estimating the circle's radius.
that was my attempt; ie I worked out the centre by plotting the points and then estimating it. helpfully the centre in this question was a whole number but I want to know how to answer the question properly(give an algebraic solution) 
 A: A middle point of $AB$ it's $(6,1)$ and $$m_{AB}=\frac{5+3}{3-9}=-\frac{4}{3}.$$
Thus, a center of the circle is placed on $$y-1=\frac{3}{4}(x-6)$$ or
$$4y-3x+14=0.$$
A: You must rember this teorem:

In a circunference, the center of the circle belongs to the perpendicular line drawn from the middle point of a chord.

So, we must calculate the middle point of $AB$, that is:
$$M(6,1)$$
Then we have to find the line passing throught $A$ and $B$, so:
$$y=-\frac{4}{3}x+15$$
Now, we can calculate the slope of the perpendicular line using $m\cdot m'=-1$, and so:
$$m'=\frac{3}{4}$$
This line must pass thriught $M$ and so, the equation becomes:
$$y=\frac{3}{4}x-\frac{7}{2}$$
Passing to the canonical form:
$$4y-3x+14=0$$
A: If $A$ and $B$ lie on a circle, then by definition they’re equidistant from its center. So, write down the standard expressions for the distance of an arbitrary point $(x,y)$ from each of these two fixed points and equate them: $$\sqrt{(x-3)^2+(y-5)^2}=\sqrt{(x-9)^2+(y+3)^2}$$ Now simplify this equation. Start by squaring both sides.
