Coordinate Geometry and Distance FormulaDate: 10/02/2002 at 16:50:01 From: Chelsey Subject: Coordinate geometry using an equation and the distance formula I need to find the center and radius of the following equations and then graph them: (x-3)^2 + (y+4)^2 - 9 = 0 x^2 + y^2 = 36 I tried to use the distance formula d= sqrt(x2-x1)^2 + (y1-y2)^2. I tried x-4 + y+16 = 9. I moved the -4 over and the 16 over and got -3. But I don't know if this is correct or if the -3 is the x or y coordinate. I used the distance formula theorem when I was given two sets of coordinates to find the radius of a circle, which I got right (that was easy). But my book doesn't explain how to find the center and radius, and graph it without being given any coordinates. I looked back in my old algebra I book but it doesn't explain anything about finding the center, radius, or coordinates. Thank you for helping me, Chelsey Date: 10/02/2002 at 17:07:35 From: Doctor Jerry Subject: Re: Coordinate geometry using an equation and the distance formula Hi Chelsey, The first equation is (x-3)^2 + (y-(-4))^2 = 3^2. You can think about the left side as the square of the distance from (x,y) to (3,-4), which according to the equation is 3^2. So the distance from (x,y) to (3,-4) is 3, and (x,y) is on a circle with center (3,-4) and radius 3. A circle with center at the point (h,k) and radius r is, using the distance formula (x-h)^2 + (y-k)^2 = r^2 This is just like my first paragraph. So, x^2 + y^2 = 36 = 6^2 is the equation of a circle with center (0,0) and radius 6. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 10/05/2002 at 20:02:28 From: Chelsey Subject: Coordinate geometry using an equation and the distance formula The equation says -3, +4, why did you change it to -(-4)? I know it is the same, but why did you do it that way? Is it because the distance formula is subtraction instead of addition? I don't see how you got coordinates of (0,0) for the second equation. Also, thanks for helping me. Chelsey Date: 10/05/2002 at 20:58:00 From: Doctor Ian Subject: Re: Coordinate geometry using an equation and the distance formula Hi Chelsey, Think about a circle with radius k centered at the origin. The equation of the circle is x^2 + y^2 = k^2 To see why this is true, you can pick any point on the circle and drop a line from there to the x-axis. You'll form a right triangle whose sides will be x and y, and whose hypotenuse will be r. The Pythagorean theorem tells us that for any right triangle, with legs A and B, and hypotenuse C, it will be true that A^2 + B^2 = C^2 In the case of a circle, the hypotenuse is always the same - it's just the radius of the circle - and any (x,y) pair that you choose on the circle satisfies the Pythagorean theorem. Does that make sense? Now, if we move the circle n units to the right, the equation of the new circle will be (x-n)^2 + y^2 = k^2 Why does the equation change this way? One way to think about it is this. Let's consider an arbitrary curve on the x-y plane: * | * | * | * | | * -----------------+-----*------- | | | * | * Suppose we move the curve to the right by one unit. This is the same as moving the y-axis to the left by one unit. * . | * . | * . | . * | . | * ------------.----+-----*------- . | . | . | * . | * Now, whatever used to be going on at (x,whatever) is going on at (x-1,whatever) instead. So using (x-1) for x everywhere is the same as shifting everything to the right by one unit. Does that make sense? Similarly, using (x+1) for x is the same as shifting everything to the left. Using (y-1) for y shifts everything up, and using (y+1) for y shifts everything down. Now, what happens if we shift a circle to the right by one unit? The center, which used to be at (0,0), is now at (1,0). In general, if we shift the circle x^2 + y^2 = k^2 to the right by n units, and up by m units, we get the equation (x-n)^2 + (y-m)^2 = k^2 So we like this form, because it's practically a description of the circle: "The radius is k, and the center is at (n,m)." Does that answer your question? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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