Question

Show that (7,2) and (1,-6) are on a circle whose center is (4,-2) and find the length of the radius

Answer 1

Do you know the standard equation for a circle?

Answer 2

x^2+y^2=r^2 ??

Answer 3

\[(x-h)^2+(y-k)^2 = r^2\]

Answer 4

Where (h, k) are the center point

Answer 5

ok with (h,k) being the center and r is the radius ..what nxt..

Answer 6

Pick a point, (7,2):

It can be odd sometimes, because the values may change sign

Answer 7

(x-4)^2+(y+2)^2=r^2

Answer 8

ok
so (3)^2+(4)^2=r^2
9+16=r^2
25=r^2
5=r
?

Answer 9

So check if the points are in the circle

Answer 10

idk how to...??

Answer 11

which point??

Answer 12

Hmm, I think I did something wrong - (7,2) is in the circle, but not (1, -6)

Answer 13

but we have to show that those points are on the circle

Answer 14

hold on

Answer 15

ok

Answer 16

sorry, I can't help you. It's been too long

Answer 17

@Nnesha

Answer 18

@campbell_st

Answer 19

ok... so the general form is

\[(x - h)^2 + (y - k)^2 = r^2\]

where (h, k) is the centre and r is the radius..

so if you substitute your centre you get

\[(x - 4)^2 + (y + 2)^2 = r^2\]

is that ok so far...?

Answer 20

yes

Answer 21

and reading the notes you have done the correct calculation

so picking the point (7, 2) and substituting that you can find r^2

\[(7 - 4)^2 + (2 + 2)^2 = r^2\]

so

\[r^2 = 25\]

so just take the square root of 25 to get the radius

Answer 22

it doesn't matter which point on the circle you use, as you'll get the same value

so that just means the equation is

\[(x - 4)^2 + (y + 2)^2 = 25\]

Answer 23

hope it makes sense

Answer 24

f you want to check you can find the distance from the centre to the points using the distance formula and show the distance is 5 units

Answer 25

but they say show that the points (7,2) and (1,-6) are on the circle... i think thats what confuses me... idk how to show this...

Answer 26

sure so you know

\[(7 - 4)^2 + (2 + 2)^2 = r^2\]

using the other point

\[(1 - 4)^2 + (-6 + 2)^2 = r^2\]

and you'll find

\[r^2 = r^2 ~~~or~~~~25 = 25\]

Answer 27

so \[\sqrt{(7-1)^2+(2+6)^2}=10\]

Answer 28

so since both points are the same distance from the centre, by definition, the points are on a circle centre (4, -2) and radius 5

Answer 29

oh \[\sqrt{(4-1)^2+(-2+6)^2}=5\]

Answer 30

I wouldn't find the distance between the points.

I'd use distance from the centre to each point... which is really the same as substituting into the general form to show that the radius is equal.

A simple definition of a circle is a set of points equidistant from a fixed point.

so showing that the radius is 5 from both points to a fixed point( the centre) then the 2 points are on a circle, centre (4, -2) and radius 5

Answer 31

and \[\sqrt{(4-7)^2+(-2-2)^2}=5\]

Answer 32

that works.... so you have shown the points are the same distance from the fixed point

Answer 33

cool thanks