Question

A circle that has (3,-1) and (7,-5) as endpoints of a diameter.
Use the given information to write the standard equation of the circle.

Answer 1

plz i need someones help

Answer 2

you need two pieces of information to write the equation for a circle
- the radius r
- the center (h,k)

standard equation is (x-h)^2 + (y-k)^2 = r^2

you're given the endpoints of the diameter. to find the radius, use the distance formula on the two points to find the diameter, and then divide that by 2 to get the radius. that'll be your r value

to find the center, use the midpoint formula on the two points to get the midpoint (h,k)

plug h, k, and r into the standard equation (x-h)^2 + (y-k)^2 = r^2

Answer 3

im rlly confused how to put it down cs my geometry teacher gets the questions wrong himself so i dont rlly know to put it down as

Answer 4

can you help me?

Answer 5

so you know the general standard equation for a circle now, (x-h)^2 + (y-k)^2 = r^2, and you need to find the h, k, and r values based on the information I've given you. are you getting stuck on some specific part?

Answer 6

all of it cs my teacher hasnt even taught us this

Answer 7

to describe a circle, the standard equation is (x-h)^2 + (y-k)^2 = r^2, where r is the radius, and (h,k) is the center. h is the x-value of the center, k is the y-value of the center.

the problem is asking you to write the equation of a circle. its diameter has endpoints (3,-1) and (7,-5).

to find the length of the diameter, we need the distance between those two points. to do that, we use the distance formula

\[d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\]
plug in your two points (3,-1) and (7,-5) into this formula to find the distance. pick one of the points to be (x1,y1) and the other one to be (x2,y2)

Answer 8

ok i did the distance formula and i got a decimal is tht correct?

Answer 9

what number did you get?

Answer 10

i got the distance formula as 6.32455532

Answer 11

I think you made some arithmetic error somewhere

\[d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\]

plugging in (3,-1) and (7,-5)

\[d=\sqrt{(7-3)^{2}+(-5-(-1))^{2}}\]=\[\sqrt{4^2+(-4)^{2}}=\sqrt{32}=\sqrt{4*4*2}=4\sqrt{2}\]

let's leave it as a radical for now.

if the diameter is 4sqrt(2), that means our radius is one half of that, our 2sqrt(2)

so that's our r value, let's move onto the center, which is thankfully a bit easier

Answer 12

how do you get
\[4\sqrt{2}\]

Answer 13

\[d=\sqrt{(7-3)^{2}+(-5-(-1))^{2}}\]=\[\sqrt{4^2+(-4)^{2}}=\sqrt{32}=\sqrt{4*4*2}=4\sqrt{2}\]

32 is equal to 4 * 4 * 2

the square root of (4 * 4) is 4

so sqrt(4 * 4 * 2) becomes 4*sqrt(2)

Answer 14

ok

Answer 15

so whts do you do next?

Answer 16

now that we have the radius r, we can find the center of the circle

the center of a circle is the midpoint of the diameter

our diameter has endpoints (3,-1) and (7,-5), so we need to find the midpoint of these two points

midpoint formula: \[(\frac{ x_{1}+x_{2} }{ 2 },\frac{ y_{1}+y_{2} }{ 2 })\]
basically, add the x-coordinates of (3,-1) and (7,-5), divide by 2, that's the x-coordinate of the midpoint

repeat with the y-coordinates to get the y-coordinate of the midpoint

lmk what you get or if you're still stuck

Answer 17

ok

Answer 18

question do you subtract y1 and y2 together or do you subtract since they are bothe a negative?

Answer 19

you add the two values, regardless of their sign

-1 + (-5) is the same as -1 - 5 = -6

Answer 20

ok so it would be -6 divided by 2?

Answer 21

yes

Answer 22

ok and after you got them added you divide them?

Answer 23

yes, so -6/2 = -3 as your y-coordinate

Answer 24

and 5 as my x- coordinate

Answer 25

so wht do i do next?

Answer 26

good, so (5,-3) is your midpoint and thus the center of your circle

putting everything together: r = 2sqrt(2), (h,k) = (5,-3)

standard equation is (x-h)^2 + (y-k)^2 = r^2

so plugging in the r, h, and k values

\[(x-5)^{2}+(y-(-3))^{2}=(2\sqrt{2})^2\]
which simplifies to
\[(x-5)^{2}+(y+3)^{2}=8\]as our standard form equation

Answer 27

THANK YOU SO MUCH!