Question

What is the center and radius of the circle with equation (x - 5)2 + (y + 3)2 = 16?

Answer 1

Do you know the standard equation of a circle?

Answer 2

no i dont...

Answer 3

ok :) Write this on a flash card!!

Standard equation of a circle is: \(\large (x-h)^2 +(y-k)^2 = r^2\)

Answer 4

lol i will! i think its one of these 2?

a) center (5, -3); radius = 4
b) center (3, -5); radius = 4

Answer 5

but idk maybe im wrong 0:

Answer 6

The center is located at the point (h,k) and the radius is just r.

Answer 7

im still lost...

Answer 8

are a and b both wrong?

Answer 9

Well, remember, uor formula states that the standard form of a circle is :\(\large (x-\color{green}h)^2 +(y-\color{green}k)^2 = \color{blue}r^2\) so our center is \(\large \color{green}{ (h,k)}\) and our radius is \(\large \color{blue}r\)

Answer 10

We'll find out if they are right in a minute xD

Answer 11

Our problem states we have the equation \(\large (x-5)^2 + (y+3)^2 = 16\) and we want to put it in standard form, just like that formula above.

Answer 12

so the answer should be center (-5, 3); radius = 16 right? :)

Answer 13

We can rewrite this equation as \(\large (x-\color{green}5)^2 +(y-\color{green}{(-3)})^2 = \color{blue}{(4)}^2\)

Answer 14

So now if we compare our rewritten equation to our standard form, what can we say our center is?

Answer 15

it would have to be: center(3, -5) ; radius = 4 ?? 0:

Answer 16

remember, our center is the point \((h,k)\) and our standard form is \((x-h)^2 + (y-k)^2 = r^2\)

Answer 17

The points cannot switch around D:

Answer 18

hmmm so center (5, -3); radius = 4 ???

Answer 19

Yes :D good job. (5,-3) corresponds to \((x-\color{red}5)^2 +(y-\color{red}{(-3)})^2= 4^2\) and our radius is the square root of 16, which gives us 4.

Answer 20

ahh thank you! :DD

Answer 21

no problemo.

Answer 22

:) good luck!

Answer 23

thanks!