1. ## Circle Geometry Help Please

A (-6,2) and B (-4,-2) are endpoints of a chord of a circle. C (2.-4) and D (8,4) are endpoints of a second chord.

a) Determine the coordinates of the center of the circle.

b) Determine the radius of the circle.

2. Hi

The center of the circle is equidistant from A and B therefore is on the mediator of [AB]
It is also on the mediator of [CD]

3. Originally Posted by PiratePrincess
A (-6,2) and B (-4,-2) are endpoints of a chord of a circle. C (2.-4) and D (8,4) are endpoints of a second chord.

a) Determine the coordinates of the center of the circle.

b) Determine the radius of the circle.

An alternative way is...

The centre of the circle is the same distance away from all 4 points.

If the centre is (x,y) then we can use the distance formula
on any 3 of the 4 points to find (x,y).

$[x-(-6)]^2+[y-2]^2=[x-(-4)]^2+[y-(-2)]^2=[x-2]^2+[y-(-4)]^2$

$(x+6)^2+(y-2)^2=(x+4)^2+(y+2)^2=(x-2)^2+(y+4)^2$

Multiply this out

$x^2+12x+36+y^2-4y+4=x^2+8x+16+y^2+4y+4=x^2-4x+4+y^2+8y+16$

Since $x^2+y^2$ is common to all three, we can eliminate them

$12x-4y+40=8x+4y+20=-4x+8y+20$

Combining the first two equations to write x in terms of y

$12x-8x+40=4y+4y+20\ \Rightarrow\ 4x+40=8y+20\ \Rightarrow\ x+10=2y+5\ \Rightarrow\ x=2y-5$

Using this in the 2nd and 3rd equations

$8(2y-5)+4y+20=-4(2y-5)+8y+20$

$16y-40+4y+20=-8y+20+8y+20$

$20y-20=40$

$20y=60\ \Rightarrow\ y=3\ \Rightarrow\ x=6-5=1$

The radius is the distance from the centre to any of the 4 points

$r=\sqrt{(8-1)^2+(4-3)^2}=\sqrt{7^2+1^2}=\sqrt{50}$

4. ## Thank you!

Thank you so, so much! I wish I could hug you!
I'm a Mom of three, taking a grade 12 math class at night. I haven't been in a classroom in over 15 years and I'm finding it very challenging (to put it nicely lol).
This forum has been invaluable to me.

5. ## circle geometry help please

Originally Posted by PiratePrincess
A (-6,2) and B (-4,-2) are endpoints of a chord of a circle. C (2.-4) and D (8,4) are endpoints of a second chord.

a) Determine the coordinates of the center of the circle.

b) Determine the radius of the circle.

running back several days I noticed this problem and took a crack at it but
unfortunately discovered that your two lines are chords of different circles

6. Originally Posted by bjhopper
running back several days I noticed this problem and took a crack at it but
unfortunately discovered that your two lines are chords of different circles
Distance from (1,3) to (-6,2) is $\sqrt{7^2+1^2}$

Distance from (1,3) to (-4,-2) is $\sqrt{5^2+5^2}$

Distance from (1,3) to (2,-4) is $\sqrt{1^2+7^2}$

Distance from (1,3) to (8,4) is $\sqrt{7^2+1^1}$

In all cases the distance is $\sqrt{50}$

hence both are chords of the same circle centred at (1,3) with radius $\sqrt{50}$

7. ## circle geometry

[quote=Archie Meade;469199]Distance from (1,3) to (-6,2) is $\sqrt{7^2+1^2}$

Distance from (1,3) to (-4,-2) is $\sqrt{5^2+5^2}$

Distance from (1,3) to (2,-4) is $\sqrt{1^2+7^2}$

Distance from (1,3) to (8,4) is $\sqrt{7^2+1^1}$

In all cases the distance is $\sqrt{50}$

hence both are chords of the same circle centred at (1,3) with radius $\sqrt{50}$[/quote

Sorry I'm wrong. Iplotted the points and got slightly different radii lenghts and although th center was 1,3 i should have calculated all the radii