How to calculate the centre of a circle given two points and the equation of a line going through the centre?

Question

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)

Answer 1

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 $$4y-3x+14=0.$$

Answer 2

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: $$4y-3x+14=0$$

Answer 3

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.