Question

Write an equation for a circle with center at (2,3) and radius of 6.

Answer 1

@Jonask

Answer 2

we need to write as
\[(x-a)^2+(y-b)^2=r^2\]
where (a,b) is center and r=6

Answer 3

can you do that

Answer 4

ok so \[(2-a)^2+(3-b)^2=6\] like that

Answer 5

substitue in the place of a and b but leave x and y

Answer 6

ummmm so \[(x-2)^2+(y-3)^2=6\] this way????

Answer 7

yes but \[6^2\] on the right

Answer 8

oh ok

Answer 9

yes no p

Answer 10

???????

Answer 11

no problem

Answer 12

Write an equation for a circle with center at (-2,7) and diameter of 14.

Answer 13

now we first find the radius
ie half of the diameter 14/2=7=r
then do the same procedure as before

Answer 14

ok so the answer is
\[(x+2)^2+(y-7)^2=14^2\]

Answer 15

yes but be care ful i explained above wat u shud do to 14

Answer 16

oh its 7^2

Answer 17

yes good

Answer 18

great job

Answer 19

this next one has a graph

Answer 20

whe we count the blocks from center we see that the radius is r=3
center(3,-1)

Answer 21

ok the answer is
\[(x-3)^2+(y+1)^2=3^2\]

Answer 22

yes good

Answer 23

ok 1 more
Use the equation below to find the center and radius of the circle.

x2 -2x +y2 -2y -23 =0
Select one:
a. (-1,-1) r=23
b. (1,-1) r=25
c. (1,1) r =5
d. (1,1) r= 23

Answer 24

\[ x^2 -2x +y^2 -2y -23 =0\]
you need to complete square

Answer 25

\[(x -1)^2+(y-1)^2=23+1+1=25 \]

Answer 26

center(1,1) r=5

Answer 27

ok thank u