Question

4x^2+4y^2-16x-24y+51=0 is the equation of a circle. where is the center of the circle and what is it's radius?

A. center (2,3) radius=1/4
B. center (2,3) radius=1/2
C. center (-2,-3) radius=1/4
D. center (-2,-3) radius=1/2

Answer 1

how u do it

Answer 2

divide by 5
then complete the square twice

Answer 3

do you know how to complete the square?

Answer 4

no

Answer 5

algebra 2 new work

Answer 6

then you can't do it

Answer 7

i dont know how to do it

Answer 8

this is not a good place to start, because this is much more complicated to complete the square
not that it is that difficult, but you would be better off starting with something like
\[x^2+6x=5\] and solving by completing the square
it takes a few minutes to learn it

Answer 9

if you have not seen it this is not a good starting point
i would cheat
http://www.wolframalpha.com/input/?i=circle+4x^2%2B4y^2-16x-24y%2B51%3D0

Answer 10

what i do now?

Answer 11

i got 4x(x-4) +4y(y-6)=-51

Answer 12

divide by 4

Answer 13

@satellite73 with what i have

Answer 14

you really don't want to factor like that

Answer 15

@satellite73 can u show me then

Answer 16

\[4x^2+4y^2-16x-24y+51=0\] divide by 4\[x^2-4x+y^2-8y=-\frac{51}{4}\]

Answer 17

then take half of 4, which is 2 and write
\\[(x-2)^2+y^2-8y=-\frac{51}{4}+2^2\] repeat with the y terms get
\[(x-2)^2+(y-4)^2=-\frac{51}{4}+2^2+4^2\]

Answer 18

AH DAMN

Answer 19

one fourth of 24 is 6 back to the drawing board

Answer 20

its not right

Answer 21

\[x^2-4x+y^2-6y=-\frac{51}{4}\\
(x-2)^2+(y-3)^2=-\frac{51}{4}+2^2+3^2\]

Answer 22

@satellite73 i get 1/4 not 1/2 idek what i did

Answer 23

@myininaya i get 1/4 what happened

Answer 24

@myininaya hello?

Answer 25

write the equation in the form
\[x^2+y^2+2gx+2fy+c=0\]
center is (-g,-f) and
\[radius=\sqrt{g^2+f^2-c}\]

Answer 26

\[ax^2+by^2+cx+dy+e=0 \\ ax^2+cx+by^2+dy=-e \\ (ax^2+cx)+(by^2+dy)=-e \\ \text{for this to be a circle we need } a=b \\ \text{ so I'm going to replace } b \text{ with } a \\ (ax^2+cx)+(ay^2+dy)=-e \\ \text{ now we are going to need the coefficients of } \\ x^2 \text{ and } y^2 \text{ inside the ( ) to be 1 } \\ \text{ but there are } a's \text{ in front of them } \\ \text{but there are no a's next to the } cx \text{ and } dy \text{ terms } \\ \text{ we can put } a \text{ there } \text{ as long as we also divide by } a\\ (ax^2+\frac{a}{a}cx)+(ay^2+\frac{a}{a}dy)=-e \\ \text{ now now we are ready to factor those } a's \text{ out of the ( ) } \\ a(x^2+\frac{1}{a}cx)+a(y^2+\frac{1}{a}dy)=-e \\ a(x^2+\frac{c}{a}x)+a(y^2+\frac{d}{a}y)=-e \]
\[\text{ So now we are going to put a ? in place of the numbers we need to go } \\ \text{ there to complete the square for the thingys in the ( )} \\ a(x^2+\frac{c}{a}x+?_1)+a(y^2+\frac{d}{a}y+?_2)=-e+a \cdot ?_1 +a \cdot ?_2 \\ \text{ I added } a \cdot ?_1 \text{ and } a \cdot ?_2 \text{ on both sides } \\ \text{ Now if we recall to complete the square we need } x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]

\[\text{ so let's change these ?'s to the numbers we need to write the thingys } \\ \text{ in () has something squared } \]
\[\text{ Now we are going to look at the numbers in front of } x \text{ and } y \\ \text{ take those numbers divide by 2 then square } \\ \text{ and that is what we will add in } \]

\[a(x^2+\frac{c}{a}x+(\frac{c}{2a})^2)+a(y^2+\frac{d}{a}y+(\frac{d}{2a})^2)=-e+a \cdot (\frac{c}{2a})^2+a \cdot (\frac{d}{2a})^2 \\

\text{ So now we are ready to write as } ()^2 \\ a(x+\frac{c}{2a})^2+a(y+\frac{d}{2a})^2=-e+a \frac{c^2}{4a^2}+a \frac{d^2}{4a^2}\]
\[\text{ We also need to divide both sides by } a \]
\[(x+\frac{c}{2a})^2+(y+\frac{d}{2a})^2=-\frac{e}{a}+\frac{c^2}{4a^2}+\frac{d^2}{4a^2}\\
\text{ Let's see if we can clean this up a little } \\ (x+\frac{c}{2a})^2+(x+\frac{d}{2a})^2=\frac{-e(4a)}{a(4a)}+\frac{c^2}{4a^2}+\frac{d^2}{4a^2} \\ (x+\frac{c}{2a})^2+(x+\frac{d}{2a})^2=\frac{-4ae+c^2+d^2}{4a^2}\]

\[\text{ Checking your answer } \\ \text{ Center is } (-\frac{c}{2a},-\frac{d}{2a}) \\ \text{ and the radius is } \sqrt{\frac{-4ae+c^2+d^2}{4a^2}}\]
\[\text{ you had } \\ a=4 \\ b=4 \\ c=-16 \\ d=-24 \\ e=51 \\ (x+\frac{-16}{2(4)})^2+(y+\frac{-24}{2(4)})^2=\frac{-4(4)(51)+(-16)^2+(-24)^2}{4(4)^2} \\ (x-2)^2+(y-3)^2=\frac{-16(51)+(16)^2+(24)^2}{4(16) } \\ (x-2)^2+(y-3)^2=\frac{16(-51)+(16)(16)+(4 \cdot 6)^2}{16(4)} \\ (x-2)^2+(y-3)^2=\frac{16(-51)+16(16)+16(6)^2}{16(4)} \\ (x-2)^2+(y-3)^2=\frac{-51+ 16+6^2}{4}\\ (x-2)^2+(y-3)^2=\frac{-51+16+36}{4} \\ (x-2)^2+(y-3)^2=\frac{-51+52}{4} \\ (x-2)^2+(y-3)^2=\frac{1}{4} \\ \text{ so yeah } \\ \text{ Center } (\frac{-c}{2a},\frac{-d}{2a})=(2,3) \text{ and radius is } \sqrt{\frac{-4(4)(51))+(-16)^2+(-24)^2}{4(4)^2}}=\sqrt{\frac{1}{4}} \]


\[\text{ So you need to just take the square root of that 1/4 }\]