Question

Please help me?***WILL MEDAL AND FAN***

Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit.

x2 −2x + y2 − 6y = 26

Answer 1

either complete the squares and write in the form
(x-h)^2+(y-k)^2=r^2
center is (h,k) and radius =r
second method
compare with \[x^2+y^2+2gx+2fy+c=0\]
center is (-g,-f) and radius \[r=\sqrt{g^2+f^2-c}\]

Answer 2

you know how to complete the square?

Answer 3

@satellite73 nope

Answer 4

@sshayer i dont understand anything you said because of the way you worded it

Answer 5

Have you read any explanations (in your book or online) of how to find the center and vertex of a circle? about how to "complete the square?" If not, now would be a good time to look up those two topics.

Answer 6

@mathmale ok thanks

Answer 7

@mathmale i sarched it up but it doesnt make any sense

Answer 8

*searched

Answer 9

chocolates!
first i'm going to put parentheses to separate x's and y's terms \[\rm (x^2 −2x )+( y^2 − 6y )= 26\] now we have to complete the square for each parentheses

Answer 10

take half of the middle term(coefficient of x term)
square it and then add this value to both side
here is an example
\[(z^2-6z)= 4\] take half of middle term (which is 6 in this example )
6/2= 3
now square this value (3)^2 = 9
add the resultto the both sides
\[(z^2-6z\color{Red}{+9})= 4\color{red}{+9}\]

Answer 11

you can factor that quadratic equation at left side
but in short form the factor of the trinomial that you get after completing the square would always be \[\rm (x-\frac{b}{2})^2\]

Answer 12

wait.. I dont under stand the (z^2- 6z part could you explain it again

Answer 13

sure. thanks for asking

Answer 14

btw thank you sooo much for helping

Answer 15

\[\large\rm Ax^2+Bx+C=0\] the quadratic equation
where `b` is the middle term ( the coefficient of x term)
first we need to see if the leading coefficient is one (if not we need to solve the equation to get rid of the leading coefficient )

to complete the square you have to take half of `b` ( coefficient of x term )

Answer 16

\[\large\rm x^2+Bx+C=0\]
for example this is my equation
1st) leading coefficient is one !
2nd) move the constant term to the right side
\[\large\rm \color{blue}{x^2+Bx} =-C\]
complete the square of blue part
3rd) take half of middle term ( b) and then square it ( add the result both side of the equation )
\[\rm x^2+Bx +\color{Red}{(\frac{b}{2})} =-C \color{Red}{+(\frac{b}{2})^2}\]
(in simple words { divide b by 2, square it , add this value both sides )

Answer 17

ohh ok

Answer 18

there is a typo

Answer 19

where

Answer 20

\[\rm x^2+Bx +\color{Red}{(\frac{b}{2})^2} =-C \color{Red}{+(\frac{b}{2})^2}\]
**

Answer 21

oh

Answer 22

Thank you!

Answer 23

\[\rm \color{red}{x^2+Bx +}\color{Red}{(\frac{b}{2})^2} =-C \color{black}{+(\frac{b}{2})^2}\]
and then factor the left side which will always be \[\rm (x+\frac{b}{2})^2 \]
(x+half of middle term )^2

Answer 24

np. can you complete the square of `x^2-2x` ?

Answer 25

I think so

Answer 26

2/2=1
1*1=1 so just add the 1 right?

Answer 27

okay go ahead give it a try!

Answer 28

yes that's correct.
btw `x^2-2x` b=-2
take half -2/2 = -1 and when you take square of negative number you will always get the positive answer. nice

Answer 29

You are a life saver!

Answer 30

what would be the next step?
write the the equation form

Answer 31

in**

Answer 32

Would i do the same thing for the other parentheses?

Answer 33

since i have two numbers to add on which do i add on the 26

Answer 34

\(\color{blue}{\text{Originally Posted by}}\) @DivinexxChocolatexx
Would i do the same thing for the other parentheses?
\(\color{blue}{\text{End of Quote}}\)
not yet
we still have to work on first parentheses
add (b/2)^2 both sides \[(\color{Red}{x^2 -2x+1}) +(y^2-6y)=26+(-1)^2\]what are the factors of this perfect square trinomial ?

Answer 35

tbh I don't know

Answer 36

wait is it 2 and 6?

Answer 37

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
\[\rm \color{red}{x^2+Bx +}\color{Red}{(\frac{b}{2})^2} =-C \color{black}{+(\frac{b}{2})^2}\]
and then factor the left side which will always be \[\rm (x+\frac{b}{2})^2 \]
(x+half of middle term )^2
\(\color{blue}{\text{End of Quote}}\)

Answer 38

x^2-2x+1
when the leading coefficient is one
then find two numbers when you multiply them you should get product of `A times C`
and the sum of same numbers should be equal to `b` ( the middle term )

Answer 39

Ax^2+Bx+C=0
quadratic equation where
a=leading coefficient
b= coefficient of x term( middle term)
c=constant

Answer 40

2 and 26?

Answer 41

first of all
x^2-2x+1
list a , b ,c values

Answer 42

a value is 2 and b value is 6 and c value is 26

Answer 43

hmm
`x^2-2x+1` i don't see any `26`

Answer 44

ohhhh

Answer 45

so c value is 1

Answer 46

\[\rm (x^2 −2x )+( y^2 − 6y )= 26\] we are completing the square of first parentheses
\[\rm (x^2 −2x +\color{Red}{(\frac{-2}{2})^2 })+( y^2 − 6y )= 26\color{Red}{+(\frac{-2}{2})^2}\] add (b/2)^2 both sides
we get \[\rm (x^2 −2x +\color{Red}{ 1 })+( y^2 − 6y )= 26\color{Red}{+1}\]

Answer 47

add 26+1 =27
\[\rm \color{blue}{(x^2 −2x +\color{Red}{ 1 })}+( y^2 − 6y )= 27 \]
now we have to factor this trinomial(quadratic equation ) so at this point we are only dealing with the blue part (quadratic equation

Answer 48

yes c=1
a=1
b=-2
multiply A times C = 1 times 1
what two numbers would you multiply to get 1( the product of AC)
and when you add these same numbers you should get (-2) (the value of b)

Answer 49

omg that makes so much sense

Answer 50

this is a totally random question but what time is it where you are

Answer 51

we don't use clocks where i live c;

Answer 52

lmao so what you use

Answer 53

at this point you have to learn how to factor quadratic equation
but for this question when we complete the square, the factor of trinomial would always be (x+half of b)^2

Answer 54

oh i know how to do that

Answer 55

\[\rm (x^2 −2x )+( y^2 − 6y )= 26\] we are completing the square of first parentheses
\[\rm (x^2 −2x +\color{Red}{(\frac{-2}{2})^2 })+( y^2 − 6y )= 26\color{Red}{+(\frac{-2}{2})^2}\] add (b/2)^2 both sides
we get \[\rm (\color{Red}{x^2 −2x +\color{Red}{ 1} })+( y^2 − 6y )= 27\]
\[\rm (\color{Red}{(x+\frac{b}{2})^2 })+( y^2 − 6y )= 27\]
replace b with its value that's it and then simplify

Answer 56

to factor it?

Answer 57

well first just replace `b` with its value

Answer 58

oh i thought i would have to times the first set of parentheses by itself

Answer 59

Thank you for helping me but i need to go to sleep it's 1:01 in the morning

Answer 60

no no.
that's kinda short cut let me explain it first
`x^2-2x+1` factor this equation
the product of leading coefficient and constant is 1
middle term is -2
two numberes are
-1 and -1
-1 times -1 = 1 ( the product of AC)
-1-1=-2 ( sum is equal to the middle term
the factor are (x-1)(x-1)
(x + 1st number )(x+2nd ) number

Answer 61

good night :)

Answer 62

the short cut i was talking about earlier is (x+(b/d))^2
\[\rm (x^2 −2x )+( y^2 − 6y )= 26\] we are completing the square of first parentheses
\[\rm (x^2 −2x +\color{Red}{(\frac{-2}{2})^2 })+( y^2 − 6y )= 26\color{Red}{+(\frac{-2}{2})^2}\] add (b/2)^2 both sides
we get \[\rm (\color{Red}{x^2 −2x +\color{Red}{ 1} })+( y^2 − 6y )= 27\]
\[\rm (\color{Red}{(x+\frac{b}{2})^2 })+( y^2 − 6y )= 27\]
replace b with its value that's it and then simplify

b =-2 plugin
\[\rm (\color{Red}{(x+\frac{-2}{2})^2 })+( y^2 − 6y )= 27\]
simplify -2/2 = -1
\[\rm (\color{Red}{x-1 })^2+( y^2 − 6y )= 27\]
same answer
you can say (x+b/d)^2 is the formula to factor trinomial when we complete the square

Answer 63

follow same steps to complete the square for 2nd parentheses! practice! and practice on your ow
the lag is real _--
alright i'm tired too good night !