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

- LynFran

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

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- anonymous

Do you know the standard equation for a circle?

- LynFran

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

- anonymous

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

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## More answers

- anonymous

Where (h, k) are the center point

- LynFran

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

- anonymous

Pick a point, (7,2):
It can be odd sometimes, because the values may change sign

- LynFran

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

- LynFran

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

- anonymous

So check if the points are in the circle

- LynFran

idk how to...??

- LynFran

which point??

- anonymous

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

- LynFran

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

- anonymous

hold on

- LynFran

ok

- anonymous

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

- LynFran

@Nnesha

- LynFran

@campbell_st

- campbell_st

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...?

- LynFran

yes

- campbell_st

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

- campbell_st

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

- campbell_st

hope it makes sense

- campbell_st

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

- LynFran

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...

- campbell_st

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

- LynFran

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

- campbell_st

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

- LynFran

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

- campbell_st

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

- LynFran

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

- campbell_st

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

- LynFran

cool thanks

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