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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

Q: What is the equation of the circle with center -4 and 3 and radius 6?

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this is impossible

cxbdfbdb

Center of circle: (6, 8) Radius of circle: 3

center 5,-3 radius 4

This makes no sence

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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

This makes no sence

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

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