 |
| ||
| |||
|
| Math Central | Quandaries & Queries |
|
Question from Ryan, a student: hello and thank you for such a wonderful service. This problem I think needs to be checked could you take a gander at it and tell me if i get it correct thanks find the center and the radius of this circle x^2+y^2=8x-2y+15=0 I came up with center -2, 1/2 and a radius of 11 3/4 |
Hi Ryan,
I didn't get the same answer as you did so I want to show you the process with a different problem.
Find the centre and the radius of this circle x2 + y2 - 6x + 12y + 6 = 0
To see the centre and the radius I want to write the circle in the form (x - a)2 + (y - b)2 = r2 for then I know that the circle has centre (a, b) and radius r.
For the moment look at (x - a)2 = x2 - 2ax + a2. In my problem the terms containing x are x2 and - 6x and comparing the x-terms, - 2ax and -6x, a must be 3. That is a is 6/2. Similarly (y - b)2 = y2 - 2by + b2 and comparing -2by to 12y, b must be -6. Here b is -12/2. So a is minus half the coefficient of x and b is minus half the coefficient of y. So here is my solution to
Find the centre and the radius of this circle x2 + y2 - 6x + 12y + 6 = 0
x2 + y2 - 6x + 12y + 6 = 0
x2- 6x + (6/2)2 + y2 + 12y + (-12/2)2 + 6 = (6/2)2 + (-12/2)2 I added a2 and b2 to both sides
(x - 3)2 + (y + 6)2 + 6 = 9 + 36
(x - 3)2 + (y + 6)2 = 9 + 36 - 6 = 39
Hence the centre is (3, -6) and the radius is √39.
Now try your problem again.
Harley
Ryan wrote back
hello I took the advice on the problem to find the center of the circle and the radious
the prob is this x^2+y^2+8x-2y+15=0
is this correct
-4,1 r=2
Hi again Ryan,
I put the equation of the circle in standard form and got
(x + 4)2 + (y - 2)2 = 2
Thus the centre is (-4, 1) but the radius is √2.
Harley
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.