Question

Find the equation of the circle whose center and radius are given.

center ( 5, 6), radius = 3

Answer 1

you don't "find" it you just "write" it
\[(x-h)^2+(y-k)^2=r^2\] put \(h=5,k=6, r=3\)

Answer 2

let me know what you get

Answer 3

wouldn't it just equal 9?

Answer 4

yes, the right side would be \(3^2=9\)
how about the left sides? what would that be?

Answer 5

it would be 0. Because if you sub in x (which is 5) and subtract h which is 5 they would just cancel eachother out. the same for the y-k

Answer 6

and 0 to the second power is 0

Answer 7

ok lets go slow because i think you may be confused about what the question is asking for

Answer 8

did you ever do those problems where it says "find the equation of the line given blah blah"?

Answer 9

and you end up with something like \(y=2x+3\) ? does that look familiar?

Answer 10

yes

Answer 11

ok so in those problems your answer was an equation that contained both an \(x\) and a \(y\)

Answer 12

this is the same idea exactly, your answer will have an \(x\) and a \(y\) in it

Answer 13

in other words, write the following
\[(x-h)^2+(y-k)^2=r^2\] leave the \(x\) and the \(y\) in the equation
only where you see a \(h\) put a \(5\) and where you see a \(k\) but a \(6\)

Answer 14

*put a \(6\)

Answer 15

o okay. so its (x-5)^2+(y-6)^2=3^2

Answer 16

correct?

Answer 17

yes it is

Answer 18

you can write 9 on the right if you like

Answer 19

now you are done, and i hope you see it was easy, but i get the feeling that it is not really clear what this means
i could try to explain if you like, if not that is fine too

Answer 20

no I understand! its just in a linear equation form. correct?

Answer 21

well it is not a linear equation because a circle is not a line

Answer 22

it is the equation of a circle
not the equation for a line
but suppose i told you i had an equation of a line, say \(y=2x+3\)
what does that mean exactly?

Answer 23

it means you have a line where y=-3/2 and im not sure how to figure out x

Answer 24

* 3/2x

Answer 25

ok lets back up a second
math teacher often say "the equation of the line \(y=2x+3\)" but to not explain what it means, and although i don't want to slow you up it is pretty clear that you are not sure wof what it means to say that \(y=2x+3\) is the equation of a line or \((x-5)^2+(y-6)^2=9\) is the equation of a circle

Answer 26

in the example of \(y=2x+3\) as the equation for a line, it means that any point on the line will fit the equation , and any point \((x,y)\) that fits the equation is on the line

for example, \((7,17)\) is on the line \(y=2x+3\) because \(17=2\times 7+3\) is true

Answer 27

but \((2,5)\) is not on the line \(y=2x+3\) because \(5=2\times 2+3\) is not true

Answer 28

okay im getting it

Answer 29

if you draw the line given by the equation \(y=2x+3\) and pick any point on the line \((a,b)\) you will see that \(b=2a+3\)

Answer 30

there are an infinite number of points on the line, and all of them will fit that equation
now as for the circle, it is the same idea
there are an infinite number of points on the circle with center \((5,6)\) and radius \(3\)

Answer 31

and any such point \((a,b)\) will make the equation
\[(a-5)^2+(b-6)^2=9\] true

Answer 32

okay so this equation is formatted specifically for circles? just like linear equations have their own format? I think I am getting it

Answer 33

for example, \((5,3)\) is a point on the circle
how do i know?
because
\[(5-5)^2+(3-6)^2=9\] is true

Answer 34

yes, there is a specific form for the equation of a circle, which comes directly from the pythagorean theorem for a right triangle

Answer 35

in your example, you had the center \((5,6)\) and the radius \(3\)

Answer 36

okay! it makes a lot more sense now! I feel like I actually know what im doing now

Answer 37

any point on the circle must be exactly 3 units from \((5,6)\) so by the distance formula, if \((x,y)\) is on the circle, it must fit the equation
\[(x-5)^2+(y-6)^2=3^2\]

Answer 38

the general form is the one i wrote at the very beginning
if the center of the circle is \((h,k)\) and the radius is \(r\) then any point \((x,y)\) on the circle must satisfy the equation
\[(x-h)^2+(y-k)^2=r^2\]

Answer 39

okay! thank you so much!

Answer 40

yw, hope it helped a little