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February 23rd, 2012, 01:49 PM   #1
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Finding equation of a circle

Quote:
 A circle with the equation: x^2 + y^2 -8x = 0 and a hyperbola with the equation: (x^2 / 9) - (y^2 / 4) = 1 intersect at points A and B. If a new circle is created with a diameter of AB, what is the equation of the new circle?
So first I solved for y^2 in the given circle's equation to substitute into the equation of the given hyperbola:

Subtract "x^2" from both sides and add 8x to both sides:
y^2 = 8x -x^2

Replace y^2 into the given hyperbola's equation:
(x^2 / 9) - ((8x - x^2) / 4) = 1

Simplify the second part in the equation:
(x^2 / 9) - 2x + (x^2 / 4) = 1

Multiply both sides by 36:
4x^2 - 72x + 9x^2 = 36

Combine like terms and set equal to 0:
13x^2 - 72x - 36 = 0

Use the quadratic formula and I got the solutions:
x = 6
and
x = -6/13

So I have 2 x-values which I assume both are intersections made by the circle and parabola, but without graphing anything, how do I find AB?

Okay, I just tested out -6/13 for both the original circle and parabolic equations, and it comes out to an imaginary number so I guess there's no such points on them.

So one of the points is (6, sqrt(12)). How do I find the other point of intersection? February 23rd, 2012, 03:14 PM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Finding equation of a circle When you substitute 6 for x into either equation you should get 2 y-values when you take the square root. If then . So, you would have a circle with center (6,0) and radius or: Tags circle, equation, finding Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post nickqqqq1 Algebra 4 April 12th, 2013 02:56 PM soulrain Algebra 7 January 5th, 2012 07:51 PM berkeleybross Algebra 4 December 8th, 2010 07:15 PM dewjr Algebra 2 November 25th, 2010 12:00 PM aschenbr.rach Calculus 1 March 7th, 2010 11:52 AM

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