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

Mathematics
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Do you know the standard equation for a circle?
x^2+y^2=r^2 ??
\[(x-h)^2+(y-k)^2 = r^2\]

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Other answers:

Where (h, k) are the center point
ok with (h,k) being the center and r is the radius ..what nxt..
Pick a point, (7,2): It can be odd sometimes, because the values may change sign
(x-4)^2+(y+2)^2=r^2
ok so (3)^2+(4)^2=r^2 9+16=r^2 25=r^2 5=r ?
So check if the points are in the circle
idk how to...??
which point??
Hmm, I think I did something wrong - (7,2) is in the circle, but not (1, -6)
but we have to show that those points are on the circle
hold on
ok
sorry, I can't help you. It's been too long
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...?
yes
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
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 \[(x - 4)^2 + (y + 2)^2 = 25\]
hope it makes sense
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
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...
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 \[r^2 = r^2 ~~~or~~~~25 = 25\]
so \[\sqrt{(7-1)^2+(2+6)^2}=10\]
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
oh \[\sqrt{(4-1)^2+(-2+6)^2}=5\]
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
and \[\sqrt{(4-7)^2+(-2-2)^2}=5\]
that works.... so you have shown the points are the same distance from the fixed point
cool thanks

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