Question

What are the coordinates of the center of the circle shown below?
x^2+y^2-12x-6y+9=0

Answer 1

\(\normalsize\color{blue}{ x^2+y^2-12x-6y+9=0 }\) can we try to do it together this time ?

Answer 2

yes i am home now sorry about that

Answer 3

It is okay

Answer 4

First, we will say \(\normalsize\color{red}{ x^2-12x+y^2-6y+9=0 }\) what have I done to the equation ?

Answer 5

you put the x's next to the x^2 and the y next to the y^2 so that we could perfect the squares?

Answer 6

Yes

Answer 7

But we can already notice a perfect square in our (red) equation, can't we? (If we can) where is it ?

Answer 8

i think it is the y^2-6y part

Answer 9

Close, I'll underline it. \(\normalsize\color{black}{ x^2-12x+\underline{ y^2-6y+9}=0 }\)

Answer 10

See ?

Answer 11

yes i see it now

Answer 12

what do you have to add to both sides, (and were do you put it) to make a `perfect square` where the `x`s are ?

Answer 13

is it 11 and i think its over the = to the right

Answer 14

This is always the rule about a perfect square.

\(\normalsize\color{black}{ x^2±bx=c }\)
\(\normalsize\color{black}{ x^2±bx\color{red}{+(b/2)^2}=c \color{red}{+(b/2)^2} }\)

Answer 15

in this case, when looking at `x² - 12 ` what is the `(b/2)²` ?

Answer 16

it is b=-12 and i got 36

Answer 17

So, you add 36 to both sides of the equation?

Answer 18

Right ?

Answer 19

Can you do that for me please ?

Answer 20

X^2-12x+9=-36

Answer 21

hey @samsan9 do u know diffrentiation ?

Answer 22

SAMSON, Hold on, you had \(\normalsize\color{blue}{ x^2-12x+y^2-6y+9=0 }\), and we said that \(\normalsize\color{blue}{ x^2-12x+\color{red}{y^2-6y+9 }=0 }\) and the red is already a perfect square.

So,
\(\normalsize\color{blue}{ x^2-12x+\color{red}{y^2-6y+9 }=0 }\) and what do you add to the `x` to make the `x`s a perfect square ?

Answer 23

@sidsiddhartha not need in differentiation.

Answer 24

is it 27?

Answer 25

u can solve problems like this easily with partial diff
let
\[f(x)=x^2+y^2-12x-6y+9\]
now diffrentiate it partially with respect to x and equate it to zero\[\frac{ \delta f }{ \delta x }=2x-12\rightarrow 2x-12=0\rightarrow x=6\]
and
\[\frac{ \delta f }{ \delta y }=2y-6\rightarrow 2y-6=0\rightarrow y=3\]
so center is \[(6,3)\]
yeah i know @SolomonZelman it's a messy way to do it but it is quite usefull for objective questions because u can calculate centre of any conics using it very fast

Answer 26

@samsan9 forget about my method follow solmol :)

Answer 27

I agree, but considering the fact that not everyone learned differentiation.

Answer 28

yeah very true

Answer 29

i am still in precalc so i don't remember learning that in high school

Answer 30

\(\normalsize\color{blue}{ x^2-12x+y^2-6y+9=0 }\)
\(\normalsize\color{blue}{ x^2-12x\color{red}{+\rm{what?} }+y^2-6y+9=0\color{red}{+\rm{what?} } }\)

Answer 31

samson, can you do this one ?

Answer 32

x^2-12x+36+y^2-6y+9=0 and then i move the the 36 to the other side so i get x^2-12x+y^2-6y+9=36

Answer 33

you can't just add 36 to one side.

Answer 34

If you add a number you add it to both sides. You got that it is 36 correctly though, so good job !

\(\normalsize\color{blue}{ x^2-12x+y^2-6y+9=0}\)
\(\normalsize\color{blue}{ x^2-12x+36+y^2-6y+9=0+36}\)
\(\normalsize\color{blue}{ x^2-12x+36+y^2-6y+9=36}\) And now, to finish it up, re-write the `X`s and the `Y`s as perfect squares.

Answer 35

i got (x^2-12x+36)(x^2-6y+9)
and then i got (-9,-3) and ( (-3,-3)

Answer 36

you wrote 2 points `(-9,-3)` and `(-3,-3)` what are these ?

Answer 37

i got those using the diamond method

Answer 38

Alright.. but can you tell me the center?

Answer 39

(6,3)

Answer 40

No you are a little bit off

Answer 41

\(\normalsize\color{blue}{ x^2-12x+36+y^2-6y+9=36}\)

\(\normalsize\color{blue}{ \underline{x^2-12x+36}+ \underline{y^2-6x+9}=36}\)

\(\normalsize\color{blue}{ \underline{(x-6)^2}+ \underline{(y-3)^2}=36}\)

No.. I am the bad guy, you're good !

Answer 42

Good work :)

Answer 43

thank you so much for helping me out this long i really appreciate it :D

Answer 44

You welcome, it is my \(\normalsize\color{blue}{ \rm{ P:} }\)\(\normalsize\color{blue}{ leasure~~ ! }\)