Question

Complete the equation of the circle centered at (−8,9) that passes through (10,2).

Answer 1

Complete the equation of the circle centered at (−8,9) that passes through (10,2).

Answer 2

ok... so we know the center of the circle, we dunno the radius though
but we know it passes through (10,2)

so notice the circle equation --> http://www.mathwarehouse.com/geometry/circle/images/equation-of-circle/example-2-equation-of-circle.png

so... what's the radius? well, the radius will be the same distance between the center and any point on the circle, that is

\(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -8}}\quad ,&{\color{blue}{ 9}})\quad &({\color{red}{ 10}}\quad ,&{\color{blue}{ 2}})
\end{array}\qquad
d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)

Answer 3

once you have the radius, just plug the values in for h, k and r, in the equation

Answer 4

that means the answer is 9

Answer 5

i did not get that

Answer 6

hmmm well... what did you get for the distance between those 2 points?

Answer 7

36+63=99

Answer 8

(100-64)+(18+81)
36+63=99

Answer 9

i am still confuse littlebed

Answer 10

\(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -8}}\quad ,&{\color{blue}{ 9}})\quad &({\color{red}{ 10}}\quad ,&{\color{blue}{ 2}})
\end{array}\qquad
d = \sqrt{({\color{red}{ 10}}-({\color{red}{ -8}}))^2 + ({\color{blue}{ 2}}-{\color{blue}{ 9}})^2}\\ \quad \\
d=\sqrt{18^2+(-7)^2}\implies d=\sqrt{324+49}\)

Answer 11

should i add 324+49

Answer 12

well, yes, that's the radius :)

Answer 13

and get the root for the answer

Answer 14

well... you won't have to get thee root, if you can, sure
however in this case is 373, and I think 373 is a prime

Answer 15

i am sorry i am confuse i really want you to show me what is the next step

Answer 16

hmmm ok.... so.... you see how the radius was found, right?

Answer 17

yes

Answer 18

okay

Answer 19

\(\bf d={\color{green}{ \sqrt{373}}}\iff radius\quad
center\quad ({\color{blue}{ -8}},{\color{red}{ 9}})\\ \quad \\
\textit{equation of a circle}=(x-{\color{blue}{ h}})^2+(y-{\color{red}{ k}})^2=r^2\\ \quad \\
\implies (x-{\color{blue}{ -8}})^2+(y-{\color{red}{ 9}})^2=({\color{green}{ \sqrt{373}}})^2\\ \quad \\\implies (x-{\color{blue}{ -8}})^2+(y-{\color{red}{ 9}})^2={\color{green}{ 373}}\)

Answer 20

ahemm so the -(-8) will turn to +8... .thus \(\large \bf \implies (x+{\color{blue}{ 8}})^2+(y-{\color{red}{ 9}})^2={\color{green}{ 373}}\)

Answer 21

that's mean the answer is 373

Answer 22

well, that's the radius once is squared, the radius itself is \(\bf \sqrt{373}\)

Answer 23

\(\bf \bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -8}}\quad ,&{\color{blue}{ 9}})\quad &({\color{red}{ 10}}\quad ,&{\color{blue}{ 2}})
\end{array}\qquad
d = \sqrt{({\color{red}{ 10}}-({\color{red}{ -8}}))^2 + ({\color{blue}{ 2}}-{\color{blue}{ 9}})^2}\\ \quad \\
d=\sqrt{18^2+(-7)^2}\implies d=\sqrt{324+49}\implies d=\sqrt{373}\iff radius\)

Answer 24

okay the radiuse is 373 what should i do after that

Answer 25

well, if you recall the equation of a circle ---> http://www.mathwarehouse.com/geometry/circle/images/equation-of-circle/example-2-equation-of-circle.png

all you'd do is plug in the values

Answer 26

plug the value where
can you show me how

Answer 27

\(\bf d={\color{green}{ \sqrt{373}}}\iff radius\quad
center\quad ({\color{blue}{ -8}},{\color{red}{ 9}})\\ \quad \\
\textit{equation of a circle}=(x-{\color{blue}{ h}})^2+(y-{\color{red}{ k}})^2=r^2\\ \quad \\
\implies (x-{\color{blue}{ -8}})^2+(y-{\color{red}{ 9}})^2=({\color{green}{ \sqrt{373}}})^2\\ \quad \\\implies (x-{\color{blue}{ -8}})^2+(y-{\color{red}{ 9}})^2={\color{green}{ 373}}\\ \quad \\
\implies (x+{\color{blue}{ 8}})^2+(y-{\color{red}{ 9}})^2={\color{green}{ 373}}\)

Answer 28

i need you help to figure out the answer

Answer 29

well... what are you after anyway?

Answer 30

i really dont now that's why i am asking you

Answer 31

?

Answer 32

Complete the \({\color{blue}{ \textit{equation of the circle}}}\) centered at (−8,9) that passes through (10,2).

Answer 33

\(\bf d={\color{green}{ \sqrt{373}}}\iff radius\quad
center\quad ({\color{blue}{ -8}},{\color{red}{ 9}})\\ \quad \\
\textit{equation of a circle}=(x-{\color{blue}{ h}})^2+(y-{\color{red}{ k}})^2=r^2\\ \quad \\
\implies (x-{\color{blue}{ -8}})^2+(y-{\color{red}{ 9}})^2=({\color{green}{ \sqrt{373}}})^2\\ \quad \\\implies (x-{\color{blue}{ -8}})^2+(y-{\color{red}{ 9}})^2={\color{green}{ 373}}\\ \quad \\
\implies
\bbox[border: 1px solid black;padding: 5px]{ (x+{\color{blue}{ 8}})^2+(y-{\color{red}{ 9}})^2={\color{green}{ 373}}}\)

Answer 34

in the box it's the answer

Answer 35

?

Answer 36

that's the equation of the circle, that has a center at (-8, 9) and passes through (10,2).

Answer 37

so

Answer 38

hmmm yes

Answer 39

yes what???

Answer 40

n the box it's the answer <-- yes

Answer 41

it's wrong

Answer 42

it show me it's wrong

Answer 43

(x+8)^2+(y-2)^2=373

Answer 44

that's how i wrot it bot it's wrong

Answer 45

(x+8)^2+(y-9)^2=373

Answer 46

do you see ?

Answer 47

\(\bf \bbox[border: 1px solid black;padding: 5px]{ (x+{\color{blue}{ 8}})^2+(y-{\color{red}{ 9}})^2={\color{green}{ 373}}}\)

Answer 48

ohh haemm... thought there was a typo anyhow

maybe you're expected to use the squared version of the radius
the equation is fine

Answer 49

\(\bf (x+{\color{blue}{ 8}})^2+(y-{\color{red}{ 9}})^2=({\color{green}{ \sqrt{373}}})^2\)

Answer 50

okay

Answer 51

thanks any way

Answer 52

yw

Answer 53

Find an equation of the circle with center (15,−16) and radius 1 in the form of (x−A)2+(y−B)2=C where A,B,C are constant.

Answer 54

do you have any idea about this one

Answer 55

same exact thing,
only difference, you're given the values
just plug them in

Answer 56

http://www.mathwarehouse.com/geometry/circle/images/equation-of-circle/example-2-equation-of-circle.png <--- notice the example

Answer 57

i could not figure oout