LynFran
  • LynFran
Show that (7,2) and (1,-6) are on a circle whose center is (4,-2) and find the length of the radius
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Do you know the standard equation for a circle?
LynFran
  • LynFran
x^2+y^2=r^2 ??
anonymous
  • anonymous
\[(x-h)^2+(y-k)^2 = r^2\]

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

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