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Mathematics 11 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

can u helpme

OpenStudy (saifoo.khan):

@Directrix are you good with circles?

OpenStudy (anonymous):

noup

OpenStudy (anonymous):

help mee

OpenStudy (anonymous):

sir

OpenStudy (anonymous):

helloo

Directrix (directrix):

@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.

OpenStudy (anonymous):

is that the asnwer

Directrix (directrix):

No. Focus on the process, not the answer.

OpenStudy (anonymous):

yeaa

OpenStudy (anonymous):

getting it kind it

OpenStudy (anonymous):

of

OpenStudy (anonymous):

The center is (0,-2)

Directrix (directrix):

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

Directrix (directrix):

r = ?

OpenStudy (anonymous):

x^2+(y+2)^2=3 is this the standerd form

OpenStudy (anonymous):

r= -2

OpenStudy (anonymous):

hello

OpenStudy (anonymous):

are u there

Directrix (directrix):

r= -2 --> Incorrect

OpenStudy (anonymous):

then how to solve it

Directrix (directrix):

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.

OpenStudy (anonymous):

2 is r

Directrix (directrix):

Correct.

OpenStudy (anonymous):

got it

OpenStudy (anonymous):

noo r is not 2

OpenStudy (anonymous):

is it 4

Directrix (directrix):

@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.

OpenStudy (anonymous):

is it 6

OpenStudy (anonymous):

2.44

Directrix (directrix):

6 = r^2. We want r which is square root of 3.

OpenStudy (anonymous):

hmm

OpenStudy (anonymous):

trying to understand

OpenStudy (anonymous):

1.7

OpenStudy (anonymous):

did not get it it too doficult

OpenStudy (anonymous):

difficult

Directrix (directrix):

You approximated square root of 3 to get 1.7. I stick with exact form but I don't know what your options are.

OpenStudy (anonymous):

yaa i got it

OpenStudy (anonymous):

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

OpenStudy (anonymous):

now 1 more sum

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