Question

Find the equation of the circle with center at (3, 2) and through the point (5, 4).

Answer 1

@jim_thompson5910

Answer 2

(x - x1)^2 + ( y - y1)^2 = r^2
where x1 and y1 is the center

Answer 3

r is the radius of the circle

Answer 4

you can find the radius using the distance formula between the two points

Answer 5

okay I'm gonna work it out

Answer 6

isn't x the same as x1

Answer 7

(h,k) is the center
r is the radius
(x,y) is the point that lies on the circle

\[\Large (x-h)^2 + (y-k)^2 = r^2\]
\[\Large (x-3)^2 + (y-2)^2 = r^2\]
\[\Large (5-3)^2 + (4-2)^2 = r^2\]

solve for r

Answer 8

this way works as well

Answer 9

2²+2²=r²
4+4=8
8x8=64

Answer 10

you should get to \[\Large r^2 = 8\] but r is NOT equal to 64

Answer 11

(something)*(something) = 8

what is that "something"?

Answer 12

4x2

Answer 13

4 times 2 is 8

but I'm asking for two factors which are the same number

Answer 14

example:

5 times 5 = 25
6 times 6 = 36
etc

Answer 15

i don't know cause 2x2=4 and 4x4=16

Answer 16

notice how 2*2 = 4 and 3*3 = 9

so whatever that number is, it's between 2 and 3

Answer 17

2.82

Answer 18

you can apply the square root to both sides to solve for r

\[\Large r^2 = 8\]

\[\Large \sqrt{r^2} = \sqrt{8}\]

\[\Large r = ??\]

Answer 19

yes roughly 2.828

Answer 20

Thanks,sorry it took so long

Answer 21

sorry it took Me so long

Answer 22

that's ok