A circle has the equation x^2+y^2+2x-2y-14=0, determine the center and radius.
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
The equation of the circle is x^2+y^2+2x-2y-14=0. Write this in the standard form of a circle with center (h, k) and radius r: (x - h)^2 + (y - k)^2 = r^2
x^2+y^2+2x-2y-14=0
=> x^2 + 2x + 1 + y^2 - 2y + 1 = 16
=> (x + 1)^2 + (y - 1)^2 = 4^2
The center of the circle is (-1, 1) and the radius is 4.
dlccanada | High School Teacher | (Level 1) eNoter
A circle has the equation x^2+y^2+2x-2y-14=0, determine the center and radius.
Let the centre of the circle (h, k) and radious is "r".The general equation of the circle
(x-h)^2 +(y-k)^2=r^2. you have to write the given equation in the general equation of the circle form
x^2+y^2+2x-2y-14=0
(x^2+2.x.1+1^2-1^2)+(y^2-2.y.1+1^2-1^1)=14
(x+1)^2-1+(y-1)^2-1=14
(x+1)^2 + (y-1)^2=14+2
(x+1)^2 + (y-1)^2=16
(x+1)^2 + (y-1)^2=4^2
there fore compare with general equation centre(-1,1) and radius is 4