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March 31st, 2018, 05:44 AM   #1
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[ASK] Circle Equation

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.
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March 31st, 2018, 09:34 AM   #2
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let $C$ be the center.

$\|C-(0,0)\|^2 = \|C - (6,0)\|^2 $

$C_x^2 + C_y^2 = (C_x-6)^2 + C_y^2$

$C_x^2 = (C_x-6)^2$

$C_x = 6-C_x$

$C_x = 3$

And clearly $C_y=4$

The radius is thus

$r = \|(3,4)\| = 5$

thus our circle is given by

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

which simplifies to

$x^2 - 6x + y^2 - 8y = 0$

i.e answer (b)
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March 31st, 2018, 11:10 AM   #3
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If (0, 0) and (6, 0) lie on the circle then the perpendicular bisector of the segment passes through the center of the circle. The line through the points (0, 0) and (6, 0) is, of course, the x-axis so the perpendicular bisector is parallel to the y-axis. Its equation is x= 3.

The center of the circle also lies on the line y= (4/3)x. The center is x= 3, y= (4/3)(3)= 4, (3, 4). The distance from (0, 0) to (3, 4) is $\displaystyle \sqrt{9+ 16}= 5$. As a check the distance from (6, 0) to (3, 4) is $\displaystyle \sqrt{(6- 3)^2+ 16}= \sqrt{9+ 16}= 5$ also.

The equation of the circle is $\displaystyle (x- 3)^2+ (y- 4)^2= x^2- 6x+ 9+ y^2- 8y+ 16= 25$ which reduces to $\displaystyle x^2+ y^2- 6x- 8y= 0$.
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March 31st, 2018, 04:42 PM   #4
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Thanks for both of you!
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