Question

Help write the standard equation of the circle with the given center that passes through the given point

Answer 1

which problem number?

Answer 2

19 I mostly need to know how to do it for 19-24 so if you can help me with 19 then hopefully I can figure the rest out

Answer 3

Ok let's start with the graph of the two points on the same xy coordinate plane

See the attached image

Answer 4

C is the center point
P is the point that lies on the edge of the circle

Answer 5

how far are the two points from each other?

Answer 6

4 points?

Answer 7

4 units apart. Yes. You can count out the spaces between them

This trick only works if they are on the same vertical line or the same horizontal line

did you want me to go over how to use the distance formula?

Answer 8

so I just use the distance formula

Answer 9

do you know how to use it?

Answer 10

d=x2-x1^2+y2-y1^2

Answer 11

If you meant to say this
\[\Large d = \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\]
then you'd be correct

Answer 12

-2--2^2+6-10^2=-100 is what I got so that makes no sense

Answer 13

-2--2+10-6^2=16

Answer 14

Let
C = (x1,y1) = (-2,6)
P = (x2,y2) = (-2,10)

Distance from P to C is...
\[\Large d = \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\]
\[\Large d = \sqrt{\left(-2-(-2)\right)^2+\left(10-6\right)^2}\]
\[\Large d = \sqrt{\left(-2+2\right)^2+\left(10-6\right)^2}\]
\[\Large d = \sqrt{\left(0\right)^2+\left(4\right)^2}\]
\[\Large d = \sqrt{0+16}\]
\[\Large d = \sqrt{16}\]
\[\Large d = 4\]

Answer 15

I forgot to square root it sorry

Answer 16

for 20. I got 1 21. I got 7.21

Answer 17

going back to #19

d = 4 so the radius of the circle is r = 4

do you know how to get the circle equation if you know the center and radius?

Answer 18

can you give me a formula

Answer 19

Circle equation
\[\Large (x-h)^2 + (y-k)^2 = r^2\]
where
(h,k) is the center
r is the radius

Answer 20

does that help? if so, then tell me what the equation is. If not, then let me know

Answer 21

can you explain

Answer 22

in the case of problem #19

center = (h,k) = (-2,6)

so h = -2 and k = 6

r = 4 was found when we found the distance from the center to the edge

so let's plug all this in

\[\Large (x-h)^2 + (y-k)^2 = r^2\]
\[\Large (x-(-2))^2 + (y-6)^2 = 4^2\]
\[\Large (x+2)^2 + (y-6)^2 = 16\]

which is the equation of the circle with center (-2,6) and radius r = 4

Answer 23

so what coordinates do i use for x and y

Answer 24

x and y are left alone. This is so you can graph it as an equation

example: y = 3x+5

Answer 25

pretty sure this is x+y=32

Answer 26

*not right

Answer 27

which problems did you do before #19?

Answer 28

20 and 21 but all i did was distance formula

Answer 29

and I didn't do any before 19 that is where im suppose to start

Answer 30

so if you were to do say #8, then

center = (2,-8)
(h,k) = (2,-8)
h = 2 and k = -8

radius: r = 9


Plug these values in


\[\Large (x-h)^2 + (y-k)^2 = r^2\]
\[\Large (x-2)^2 + (y-(-8))^2 = 9^2\]
\[\Large (x-2)^2 + (y+8)^2 = 81\]


So the final answer to #8 is
\[\Large (x-2)^2 + (y+8)^2 = 81\]

Answer 31

the last equation I wrote is the circle equation that models the curve that graphs a circle on the xy plane. In that case, the circle is centered at (2,-8) and has radius 9

Answer 32

so the final answer for 19 is x+2^2+y-6^2

Answer 33

#19 is similar to #8

but instead of giving you the radius, you have to find the radius

Answer 34

write it out how I wrote it. Don't forget the parenthesis

Answer 35

\[(X+2)^2+(y-6)^2\]

Answer 36

yes and that is equal to 16

btw you can write it out like this

`(x+2)^2 + (y-6)^2 = 16`

notice how (x+2)^2 is completely different from x+2^2

Answer 37

oh yea

Answer 38

so that is it?

Answer 39

for 19, yes

Answer 40

awesome

Answer 41

you'll do the same for 20 through 24

Answer 42

can I send you the answers to make sure they are right

Answer 43

sure

Answer 44

20. d=1 so (x-7)^2+(y- -2)^2=1

Answer 45

not sure how you got the distance to be 1. That's not correct

Answer 46

also the x-7 should be x-1
the x-(-2) should actually be x-2

Answer 47

it didn't sound correct
sq rt of (0-1)^2+(6-2)^2=17 so 4.12?

Answer 48

\[\Large d = \sqrt{17} \approx 4.12\]
is the correct distance


so the radius is \[\Large r = \sqrt{17}\] meaning that \[\Large r^2 = 17\]

Answer 49

So the correct answer for #20 is
\[\Large (x-1)^2+(y-2)^2 = 17\]

Answer 50

wouldn't R^2= sq rt of 17 so not 17 but 4.12....

Answer 51

r itself is \[\Large r = \sqrt{17}\]

Answer 52

since distance from center to edge of circle is the radius

Answer 53

square both sides
\[\Large r = \sqrt{17}\]
\[\Large r^2 = (\sqrt{17})^2\]
\[\Large r^2 = 17\]

Answer 54

oh ok I got ya

Answer 55

so 20. (x-1)^2+(y-2)^2=17

Answer 56

yep

Answer 57

for 21. (x-7)^2+(y- -2)^2=52

Answer 58

22. (x-10)^2+y- -5)^2=125

Answer 59

for #21, make sure you write it as (y - (-2))^2 which simplifies into (y+2)^2

Answer 60

so the final answer for #21 is (x-7)^2+(y+2)^2 = 52

Answer 61

23. (x-6)^2+(y-5)^2=61

Answer 62

#22 is incorrect

Answer 63

#23 is correct

Answer 64

22 is incorrect but the others are right?

Answer 65

21 was incorrect, but I posted the answer of what it should be (scroll up)

Answer 66

oh so 22. (x+10)^2+(y+5)^2

Answer 67

and that's equal to what?

Answer 68

=125?

Answer 69

yep so the final answer for #22 is
(x+10)^2+(y+5)^2 = 125

Answer 70

24.(x+1)^2+(y+4)^2=25

Answer 71

#24 is correct

Answer 72

nice work

Answer 73

thank you so very much you were great help

Answer 74

I'm glad I could help and that it makes more sense now

Answer 75

Yea I'm reviewing problems for finals