the general equation of the circle is
(x-a)^2+(y-b)^2=r^2
where (a,b) is the center of the circle.
r is the radius of the circle
substituting the given values,
(x+4)^2+(y-3)^2=6^2
x^2+y^2+8x-6y+16+9=36
x^2+y^2+8x-6y=11
this is impossible
cxbdfbdb
Center of circle: (6, 8) Radius of circle: 3
center 5,-3 radius 4
This makes no sence
this is impossible
Centre of the circle: (3, 8) Radius of the circle: 2 Cartesian equation of the circle: (x-3)^2 + (y-8)^2 = 4
cxbdfbdb
Center of circle: (6, 8) Radius of circle: 3
center 5,-3 radius 4
If you mean a circle center at (3, 1) and a radius of 2 then the equation of the circle is (x-3)^2 +(y-1)^2 = 4
This makes no sence
The equation of a circle with center (2, -3) and radius 3 can be written as (x - 2)^2 + (y + 3)^2 = 3^2.
If you mean a circle center at (3, 1) and a radius of 2 then the equation of the circle is (x-3)^2 +(y-1)^2 = 4
(x-3)2+y2=144
I think it is center: (-4, 3) ; radius: 2 Apex:)
(x+3)2 + (y-5)2 = 9