Question

Please help!!
posting pic below!!

Answer 1

@Luigi0210

Answer 2

@Nnesha

Answer 3

@Michele_Laino can you help me???

Answer 4

please ive been stuck on this :/

Answer 5

hint:
as we can see, from the graph, the center of the circle, is:
\[\Large C = \left( { - 1.5,1.5} \right)\]
so if we consider any generic point \(P=(x,y)\) of the circle, the distance \(PC\), is:
\[\Large d\left( {C,P} \right) = \sqrt {{{\left( {x - \left( { - 1.5} \right)} \right)}^2} + {{\left( {y - 1.5} \right)}^2}} \]

Answer 6

So C?

Answer 7

hint:
of course, such distance, has to be equal to the radius of the circle, and the radius of the circle is \(r=3/2=...\)

Answer 8

that's right!

Answer 9

Choose the correct standard form of the equation of a circle given the following radius and coordinates.

r = 11, (0, 0)
Select one:
a. x^2 + y^2 = 121
b. (x – 0)^2 + (y – 11)^2 = 0
c. Either A or B
d. None of the above

Answer 10

I confirm option C

Answer 11

can you help me with the question i posted above ?

Answer 12

since the center is \(C=(0,0)\), and the radius is \(r=11\), then the distance \(d\) of a generic point of the circle from the center, is:
\[d\left( {C,P} \right) = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2}} \]
again such distance has to be equal to the radius of the circle

Answer 13

So theres more than one option? so C?

Answer 14

I think that there is only one correct option

Answer 15

I think B

Answer 16

hint:
from my replies above, we can write this:
\[\sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2}} = 11\]

Answer 17

yeah B

Answer 18

hint:
if I take the square of both sides of last equation, I get:
\[{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {\left( {11} \right)^2}\]

Answer 19

D?

Answer 20

I think it is option A, since we can write this:
\[\begin{gathered}
{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {x^2} + {y^2} \hfill \\
{\left( {11} \right)^2} = 121 \hfill \\
\end{gathered} \]

Answer 21

oh okay

Answer 22

Choose the correct standard form of the equation of a circle given the following radius and coordinates.

r = 4, (–2, 8)
Select one:
a. x2 + y2 = 4
b. (x – 2)^2 + (y – 8)^2 = 16
c. (x + 2)^2 + (y + 8)^2 = 4
d. (x + 2)^2 + (y – 8)^2 = 16

Answer 23

again, we can write this equation for the distance between a generic point \(P=(x,y)\) and the center of the circle \(C=(-2,8)\):
\[d\left( {C,P} \right) = \sqrt {{{\left( {x - \left( { - 2} \right)} \right)}^2} + {{\left( {y - 8} \right)}^2}} = 4\]

Answer 24

now, please take the square of such equation, namely:
\[{\left( {\sqrt {{{\left( {x - \left( { - 2} \right)} \right)}^2} + {{\left( {y - 8} \right)}^2}} } \right)^2} = {4^2}\]

Answer 25

C?

Answer 26

no, I'm sorry, since \(4^2=16\)

Answer 27

hint:
after a simplification, we get:
\[{\left( {x - \left( { - 2} \right)} \right)^2} + {\left( {y - 8} \right)^2} = 16\]

Answer 28

D or B

Answer 29

B

Answer 30

yes! So, option D or option B?

Answer 31

B right??

Answer 32

no, I'm sorry, option B is wrong, since in my equation we have:
\[x - \left( { - 2} \right) = x + 2\]

Answer 33

oh okay i got my signs messed up

Answer 34

would you mind?

Answer 35

if not its ok

Answer 36

I think that the point \((1,0)\) is the center, right?

Answer 37

i believe so

Answer 38

so, we can write this equation for the distance between a generic point \(P=(x,y)\) from the center:
\[\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 0} \right)}^2}} = {\left( {\frac{1}{4}} \right)^2}\]

Answer 39

and that leaves us with either A or C i believe

Answer 40

oops.. I have made a typo:
\[\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 0} \right)}^2}} = \frac{1}{4}\]

Answer 41

hint:
again, if I take the square of both sides of such equation, I can write:
\[{\left( {x - 1} \right)^2} + {\left( {y - 0} \right)^2} = {\left( {\frac{1}{4}} \right)^2}\]

Answer 42

oh no B then

Answer 43

yes! that's right!

Answer 44

thanks so much (:

Answer 45

:)