Best Answer

Equation is:

(x - -1)² + (y - 1)² = 1²

→ (x + 1)² + (y - 1)² = 1

By expanding the brackets and rearranging, this becomes:

(x² + 2x + 1) + (y² - 2y + 1) = 1

→ x² + 2x + y² - 2y + 1 = 0

Q: What is the equation for a circle with a center of negative 1 1 and a radius of 1?

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The standard equation of a circle, with center in (a,b) and radius r, is: (x-a)2 + (y-b)2 = r2

depends on the equation.

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

this is impossible

cxbdfbdb

Related questions

Equation of any circle, with any radius, and its center at any point: [ x - (x-coordinate of the center) ]2 + [ y - (y-coordinate of the center) ]2 = (radius of the circle)2

depends on the equation.

The standard equation of a circle, with center in (a,b) and radius r, is: (x-a)2 + (y-b)2 = r2

The equation of circle is (x−h)^2+(y−k)^2 = r^2, where h,k is the center of circle and r is the radius of circle. so, according to question center is origin and radius is 10, therefore, equation of circle is x^2 + y^2 = 100

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

this is impossible

cxbdfbdb

The equation of the circle is: x^2 + y^2 = 81

Centre of the circle: (3, 8) Radius of the circle: 2 Cartesian equation of the circle: (x-3)^2 + (y-8)^2 = 4

x2 + y2 = 64

x² + y² = 81.

x² + y² = 4.