More Questions on Finding the Equation of a Circle
Date: 10/1/95 at 22:35:3 From: Badar Khushnood Subject: Again in need of HELP !!! Dr. Math, I am very thankful to you for helping me out. I really appreciate your effort in this regard. I again have a question. I can solve it, but I don't know the logic behind it. If you could please solve it fully and explain all of its steps I would highly appreciate it. Ques.: Find the equation of the circle containing the points (-1, -2) and (6,-1) and touching the x-axis. Also, please write if you can provide any other sort of help. Thanking you in advance. BADAR KHUSHNOOD. LAHORE, PAKISTAN.
Date: 10/2/95 at 1:0:38 From: Doctor Andrew Subject: Re: Again in need of HELP !!! The center of the circle would be the point that is equidistant from the two points and the x-axis. You could set up an equality for the distance from a point (x,y) to each point, and an equality from the distance to (x,y) to the x-axis and each of the points. I would approach it more intuitively instead. What are the set of points that are equidistant from (-1,-2) and (6,-1)? It's a line between them, right? In fact it's the line that passes through the midpoint of the segment from (-1,-2) to (6,-1) and with a slope perpendicular to that segment. You might want to draw it just to make sure that this is the set of equidistant points from the two points. So the center of the circle has to be on this line, right? Well you can find the equation for this line since you have a slope and a point. The distance from the x-axis is just the absolute value of the y coordinate of a point. So you want to find a point that satisfies the line equation and where the distance from one of these points is which equals that absolute value. The line equation allows you to write the y-coordinate in terms of the x- coordinate. You can use the Pythagorean Theorem to find the distance from one of the two points to the point on the line. Hope this helps. -Doctor Andrew, The Geometry Forum
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