Find the center and the radius of the circle x2 + y2 – 2x + 2y + 2 = 16?

Answer 1

x^2 - 2x + y^2 + 2y + 2 = 16

x^2 - 2x + 1 + y^2 + 2y + 1 = 16
(x - 1)^2 + (y+1)^2 = 16

Center = (1, -1)
Radius = sqrt(16) = 4

Answer 2

write the above equation into the form : (x-h)^2 + (y-k)^2 = r^2 where r is the radius and (h,k) is the center point of the circle

x^2 - 2x + 1 + y^2 + 2y + 1 = 16
(x-1)^2 + (y+1)^2 = (4)^2
so r=4 and the center C(1,-1)

Answer 3

There must be 0, one, 2 or infinite units of strategies. 0 = the circles do not intersect (one thoroughly interior yet another, or totally separated) One = the circles are tangent to a minimum of one yet another 2 = commonplace intersection infinite = the circles are comparable For this difficulty, the intersection might nicely be discovered as following. i) Subtract the two equations: 12x + 6y + 24 = 0 ==> y = -2x-4 ii) replace y=-2x-4 into the two equation to acquire a quadratic equation. resolve the equation to upward push as much as two x roots.

Answer 4

if the equation of the circle isx^2+y^2+2gx+2fy+c=0 then
cenre is (-g,-f)
=(1,-1)
radius=rt(g^2+f^2-c)
rt(1+1+14)=rt16=4 units

Answer 5

0x + 4y + 2 = 16 ???

who the he** try to confuse anyone by changing the position of the number and letters!!!

x2 - 2x = 2x - 2x 'cause x2 = 2x

This is a very strange form to ask a math question!?

Answer 6

take this in to the form (x-a)2 + (y-b)2 = r2

(x-1)2 + (y + 1)2 = 16

therefore centre is (1,1) and rad = sqrt(16) = 4 units