Question

Complete the square to find the standard form of the given equation. Then give the center and radius of the circle.

X^2+y^2+16x-10y+8=0

Answer 1

This is how you complete the square : (x²+2*8x+8²) - 8² + (y²-2*5y+5²) - 5² = 0
Tell me if you need further help.

Answer 2

is that the final answer?

Answer 3

No, it's not. You need to write it in form (x-a)²+(y-b)²=r² with (a,b) being the center and r the radius of the circle.

Answer 4

can you please show me how to do that

Answer 5

I am completely lost when it comes to these sort of problems

Answer 6

Ok, do you know this rule about quadratics a²+2ab+b²=(a+b)² ?

Answer 7

yes

Answer 8

Can you apply it for (x²+2*8x+8²) and (y²-2*5y+5²) ?

Answer 9

If you could give me the answer to the problem I will go over how to do it with you as soon as I submit the answer, it is due at 9:30 please help!!

Answer 10

We'll get it done before 9:30 I can promise you that.

Answer 11

okay thank you

Answer 12

(x²+2*8x+8²)=(x+8)² and (y²-2*5y+5²) = (y-5)²
The standard form will be (x+8)²+(y-5)²=89
Hence (-8,5) the center and √89 the radius.

Answer 13

the answer says that everything is right except for the diameter and radius

Answer 14

Did you put the square root before 89 ?

Answer 15

√89=9.43

Answer 16

Im on webwork so you type it in like sqrt89

Answer 17

but if the diameter is wrong the radius will be too

Answer 18

Why did you get the diameter wrong ?

Answer 19

I didnt. They are saying that 89 is the wrong answer

Answer 20

should be 81

Answer 21

\[x ^{2}+16x+y ^{2}^{-10y=-8}\]

Answer 22

81 is right thank you

Answer 23

complete the square

\[(x ^{2}+16+64)+(y ^{2}-10y+25)=-8+64+25\]

Answer 24

Of course it is I forgot the +8 -.-

Answer 25

simplify and factor
\[(x+8)^{2}+(y-5)^{2}=81\]

Answer 26

what would you radius be of a diameter with endpoints (4,5) (12,7) and a center (8,6)?

Answer 27

the equation i have so far in standard form is (x-8)^2 + (y-6)^2 =?

Answer 28

Use the Distance Formula : d=sqrt( (x2-x1)² + (y2-y1)² ) between the points (12,7) and (8,6).
'd' will be the radius.

Answer 29

Heres another way that might help. Begin by grouping the x and y terms together inside parenthesis, like this:\[(x ^{2}+16x)+(y ^{2}-10y)=-8\]and moving the 8 over to the other side.

Answer 30

Now we can complete the square on the x terms and the y terms by taking half the x coefficient and squaring it and adding it inside the x parenthesis, and then adding it to the -8. Do the same with the y terms, to get this:\[(x ^{2}+16x+64)+(y ^{2}-10y+25)=-8+64+25\]

Answer 31

Now you can make your perfect square binomials:\[(x+8)^{2}+(y-5)^{2}=81\]This looks like a circle to me with a center of (-8,5) and a radius of 9.