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Show that the equation represents a circle by rewriting it in standard form, and find the center and radius of the circle. x2 + y2 + 4y + 2 = 0
can u helpme
@Directrix are you good with circles?
@Shaks I'm hoping you understand the process of completing the square. We'll need that to get x2 + y2 + 4y + 2 = 0 into the standard form of the equation of a circle. That standard form is (x - h) ^2 + (y - k) ^2 = r^2 where (h,k) are the coordinates of the center of the circle and r is the radius of the circle. Check out the attachment.
is that the asnwer
No. Focus on the process, not the answer.
getting it kind it
The center is (0,-2)
So, x^2 + y^2 + 4y + 2 = 0 can be written as x^2 + y^2 + 4y + __4_ = 0 + 4 (x + 0) ^2 + (y + 2) ^2 = 4 ---> Your Circle Now, look at the standard form for the circle equation, compare it to the equation, and pick out the coordinates of the center and the value of the radius. Post what you get, and I'll check. @Shaks
r = ?
x^2+(y+2)^2=3 is this the standerd form
are u there
r= -2 --> Incorrect
then how to solve it
Look at standard form for circle equation: That standard form is (x - h) ^2 + (y - k) ^2 = r^2 where (h,k) are the coordinates of the center of the circle and r is the radius of the circle. What positive number times itself = 4? That's your r.
2 is r
noo r is not 2
is it 4
@Shaks I messed up by dropping the 2. So, x^2 + y^2 + 4y + 2 = 0 can be written as x^2 + y^2 + 4y + __4_ = 2 + 4 = 6. r^2 = 6 and r is square root of 6.
is it 6
6 = r^2. We want r which is square root of 3.
trying to understand
did not get it it too doficult
You approximated square root of 3 to get 1.7. I stick with exact form but I don't know what your options are.
yaa i got it
Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.) x + 42 = x
now 1 more sum