help...anal geom?

Question

find the equation of the 0 that is concentric with x2+y2-4x-6y+4=0 and having radius of 5....
explain it please ...thanks

Answer 1

What's anal geometry? The things kids are taught nowadays!

Answer 2

The general form of the equation of a circle with a radius r and center at (a,b) is (x-a)^2 + (y-b)^2 = r^2
Completing the square:
x^2 -4x + 4 + y^2 - 6y + 9 -9 = 0
(x-2)^2 + (y - 3)^2 = 3^2
is a circle of radius 3 and center (2,3)
For a concentric circle of radius 5 use the same cinter:
(x - 2)^2 + (y - 3)^2 = 5^2

Answer 3

First we factor that equation so it will be obvious where the center is:
x²+y²-4x-6y+4=0
x²-4x+4+y²-6y+9=9
(x-2)²+(y-3)²=9
So the center of this circle is (2, 3). The radius is √9, which is 3. We need to find the circle with the radius of 5. This is simply:
(x-2)²+(y-3)²=25
Expanding:
x²-4x+4+y²-6y+9=25
x²+y²-4x-6y-12=0

Answer 4

rewrite the equation as :
(x-2)^2 + (y-3)^2 -9 = 0
this is a circle with radius sqrt(9)=3 and origin : x=2,y=3

so the circle with radius 5 and concentric \( means same origin) is

(x-2)^2 + (y-3)^2 - 25 = 0;