find the center and radius of the circle with the given equation x^2+y^2+6y=-50-14x?

Question

please help me

Answer 1

You need to transform the equation into the standard form of the equation of a circle: (x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center and r is the radius. This is done by using the process of completing the square. In your problem, rewrite the equation to be

x^2 + 14x + y^2 + 6y = -50. Now complete the square on the x terms and y terms.

x^2 + 14x + 49 + y^2 + 6y + 9 = -50 + 49 + 9
(x + 7)^2 + (y +3)^2 = 8
Now the equation is in standard form and you get the center (-7, -3) and radius sqrt(8) = 2*sqrt(2)

Answer 2

Getting the wide type through itself, x^2 + y^2 + 14*x + 6* y = -50 we desire an equation of the type (x - a)^2 + (y - b)^2 = r^2 the position the centre is (a,b) and the radius is r. (x + 7)^2 + (y + 3)^2 = [x^2 + y^2 + 14*x + 6*y] + fifty 8 = -50 + fifty 8 = 8 ...follows from increasing the above. The products enclosed in [] are the first equation. (Why the 7 after x and the three after y? To get 14*x and six*y as in accordance to question) therefore the circle has centre -7,-3 and has radius sqrt(8) <<<

Answer 3

rewrite the equation as x^2+y^2+14x+6y+50=0
centre=-14/2,-6/2=(-7,-3)
radius (g^2+f^2-c)^1/2
=(49+9-50)^1/2=(8)^1/2 units

Answer 4

12

Answer 5

center has coordinates (-7,-3) and the radius is equal to square root of 8 which is 2.828427125...

Answer 6

Call your math teacher,OR,take better notes

Answer 7

umm

Answer 8

Nice try
You do it yourself...

Answer 9

Sorry - no math! auuuugh!!