Question – Find the equation of the circle which passes through $(3, 3)$ and $(5, 7)$ and has its center on the line: a): $x-y=5$
My attempt: $$m=2, b=-3, y=2x-3$$
Then I solved equation and got $(-2,-7)$ (which I feel is odd)
$\sqrt{(3-(-2))^2+(3-(-7))^2}$
$R= 5\sqrt5$
$(x+2)^2+(y+7)^2=225$
I know this is probably wrong but can anyone help me find my mistake.
Best Answer
Hint:
The center must be on the perpendicular bisector of $(3,3), (5,7)$ which is a line segment that passes through $(4,5)$.
Further, since the slope of the line that passes through the points $(3,3), (5,7)$ is $(2)$, the slope of the perpendicular bisector must be $(-1/2)$. Therefore, you have the slope of the perpendicular bisector, and one of its points, namely the point $(4,5)$.
This implies that you can deduce the equation that represents the perpendicular bisector. You also have the equation of the second line. Therefore, you can find the intersection point of the two lines.