Question

Fan + Medal

Answer 1

general form of a circle is
\[x^2+y^2+2gx+2fy+c=0\]
and for this form
centre=(-g,-f)
\[radius=\sqrt{g^2+f^2-c}\]

Answer 2

Can you walk my through this step by step?

Answer 3

ok the equation given is-\[x^2+y^2-x-2y-\frac{ 11 }{ 4 }=0\\and~the~standard\ equation\ is \\x^2+y^2+2gx+2gy+c=0\\now \ just\ compare\ them\]

Answer 4

Where does g come in?

Answer 5

u'll get\[2g=-1\\g=-1/2\\and\\2f=-2\\f=-1\\so \ center=(-g,-f)=(1/2,1)\\radius=\sqrt{g^2+f^2-c}=\sqrt{1/4+1+11/4}=\sqrt{4}=2\]

Answer 6

g and f are just the variable terms

Answer 7

So it has to be A or B?

Answer 8

for cases like these u just have to compare with with the above standard equation
or another process is available ,that is to take it addition of square terms
which will look lke\[(x-h)^2+(y-k)^2=r^2\]
u can use any one of them

Answer 9

I literally see no similarites this is really hard for me

Answer 10

this is not hard you can try with making squares

Answer 11

Wait when did it become a square?

Answer 12

\[x^2+y^2-x-2y-11/4=0\\ [x^2-2.\frac{ 1 }{ 2 }x+(\frac{ 1 }{ 2 })^2]+[y^2-2.y+1^2]-\frac{ 11 }{ 4 }-\frac{ 1 }{ 4 }-1=0\\(x-\frac{ 1 }{ 2 })^2+(y-1)^2=\frac{ 1 6}{ 4 }=4\]
getting this?
just made two whole squared terms

Answer 13

Okay so how do I get the answer

Answer 14

so
it is of the form
\[(x-h)^2+(y-k)^2=r^2\]
compare and u'll find h and k right?

Answer 15

\[(x-\frac{ 1 }{ 2 })^2+(y-1)^2=4=2^2............(1)\\(x-h)^2+(y-k)^2=r^2.................(2)\\so~just~compare~them\]

Answer 16

h=1/2
k=1
r=2
seems okay?

Answer 17

so center=(1/2,1)
radius=2 units
same answer as before

Answer 18

So it's D or E?

Answer 19

wait it's E

Answer 20

obviously D