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 March 31st, 2018, 05:44 AM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry [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.
 March 31st, 2018, 09:34 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390 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)
 March 31st, 2018, 11:10 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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 $\displaystyle x^2+ y^2- 6x- 8y= 0$.
 March 31st, 2018, 04:42 PM #4 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Thanks for both of you!

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