Circle Equation: Solving for x^2 + y^2 + Ax + By = 0 w/y=4/3x

[Monoxdifly]

A circle L is going through the point O (0, 0) and P (6, 0). The center is in the line \(\displaystyle y=\frac{4}{3}x\). The equation of the circle L is ...
A. \(\displaystyle x^2+y^2+6x-8y=0\)
B. \(\displaystyle x^2+y^2-6x-8y=0\)
C. \(\displaystyle x^2+y^2-8x-6y=0\)
D. \(\displaystyle x^2+y^2+8x+6y=0\)
E. \(\displaystyle x^2+y^2-4x-3y=0\)

Since the equation of a circle is \(\displaystyle x^2+y^2+Ax+By+C=0\), I substituted both known points to the equation and got C = 0 as well as B = -6, so the answer is obviously B. But then my student asked "What if all options have -6 as their B? How would we know the answer?". I think it has something to do with that \(\displaystyle y=\frac{4}{3}x\), but how? Please give me some hints.

 

[MarkFL]

I think I would consider the standard form for a circle:

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

Using the given points on the circle, we then have:

\(\displaystyle h^2+k^2=r^2\)

\(\displaystyle (6-h)^2+k^2=r^2\implies 36-12h+h^2+k^2=r^2\)

Subtracting the former from the latter, we obtain

\(\displaystyle 36-12h=0\implies h=3\)

Now, we also know:

\(\displaystyle k=\frac{4}{3}h\implies k=4\)

And so this then implies:

\(\displaystyle r^2=3^2+4^2=25\)

And so putting it all together, we have:

\(\displaystyle (x-3)^2+(y-4)^2=25\)

Or:

\(\displaystyle x^2-6x+y^2-8y=0\)

 

[Monoxdifly]

Wow, that's cool! Thank you!