Question

A circle is defined by the equation x2 + y2 – x – 2y – = 0. What are the coordinates for the center of the circle and the length of the radius?

Answer 1

what the name of the paper

Answer 2

Complete the sqaures

Answer 3

\[x^2 + y^2 -x -2 y=0\\
x^2 -x +\frac 1 4-\frac 1 4+y^2 -2y +1 -1=0\]

Answer 4

K let me think a min

Answer 5

You can finish it now

Answer 6

is there opeshions

Answer 7

It seems that he left

Answer 8

k

Answer 9

you must complete the square so
for example
\[x^{2}+2x=(x+\frac{ 2 }{ 2 })^{2}- (\frac{ 2 }{ 2 })^{2}\]
completing the square for an expression ax^2 + bx
you have
\[ax^{2}+bx=a(x^{2}+\frac{ b }{ a }x)=a((x+\frac{ b }{ 2a })^{2}-(\frac{ b }{ 2a })^{2})\]

so where you have
x2 + y2 – x – 2y – = 0.
pair up the expressions of x and y

x2 -x +y2 -2y=0

then we complete the square for the x terms

x2-x
a=1
b=-1

so substitute that into what i gave you

then you complete the square for the y terms

y2-2y
a=1
b=-2

and then you make it all equal to 0

then you will move your constant to the right side

\[(x+a)^{2}+(y+b)^{2}=r^{2}\]
with a centre (-a,-b), radius r

Answer 10

\[

x^2 + y^2 -x -2 y=0\\
x^2 -x +\frac 1 4-\frac 1 4+y^2 -2y +1 -1=0\\
(x-1/2)^2 + (y-1)^2= 1/4 +1
\]

Answer 11

You are done now

Answer 12

the answer is 2 units If x² + ax + y² + by + c = 0 then (x + ½a)² + (y + ½b)² = ¼(a² + b² - 4c)
So, for the general form, the center is (-½a, -½b) and the radius is ½√(a² + b² - 4c)
Here a=-1, b=-2, and c=-11/4, so the center is (½, 1) and the radius is ½√(1 + 4 + 11) which is 2 units, hence the

Answer 13

did that help