Circle problems

[kathryn.elizabeth531]

Hi,
So, I have a few questions about circles in geometry.
There are 3 points given on the circle.
p = (1,9)
q = (4,8)
r = (6,4)

I need to find:
The coordinates of the centre of the circle.
The radius of the circle.
The equation of the circle.

I already know the answers.
The centre is (1,4)
The radius is 5.
The equation of the circle is (x-1)^2 + (x-4)^2 = 5^2

I just am not sure of theorums and equations and the work to get the answers.

Thank you.

 

[Loren]

The perpendicular bisector of pq and the perpendicular bisector of qr meet at the center of the circle. Once you have the location of the center, you have it made.

 

[garf]

if you follow the way of loren.
midpoint (p+q)/2 = (5/2 , 17/2) (p+r)/2 = (7/2,13/2)

slope m [sub:2c92a25b]pq[/sub:2c92a25b]=(-1/3)
m [sub:2c92a25b]pr[/sub:2c92a25b]=-1

perpendicular slope m [sub:2c92a25b]pq[/sub:2c92a25b]=(3)
m [sub:2c92a25b]pr[/sub:2c92a25b]=1

equation of the straight line y-(17/2) = 3(x-(5/2))

the equation of the other straight line y-(13/2) = x-(7/2)

to solve the system of equations y=3x+1 y=x+3

center =(1,4)

r=((1-1)[sup:2c92a25b]2[/sup:2c92a25b]+(9-4)[sup:2c92a25b]2[/sup:2c92a25b]) [sup:2c92a25b]1/2[/sup:2c92a25b]

 

[soroban]

Hello, kathryn.elizabeth531!

Here's another approach.
(It's longer, but it's good to have an extra weapon in your arsenal.)

There are 3 points given on the circle: .\(\displaystyle P(1,9),\;Q(4,8),\;R(6,4)\)

Find: the center, radius, and the equation of the circle.

\(\displaystyle \text{Let }C(x,y)\text{ be the center of the circle.}\)

\(\displaystyle \text{Then the distances: }\,CP,\:CQ,\:CR\text{ are all equal to the radius, }r.\)


\(\displaystyle \text{So we have: }\:\begin{array}{ccccccc}CP^2 &=&(x-1)^2 + (y-9)^2 &=& r^2 & [1] \\ CQ^2 &=& (x-4)^2 + (y-8)^2 &=& r^2 & [2] \\ CR^2 &=& (x-6)^2+(y-4)^2 &=& r^2 & [3] \end{array}\)


\(\displaystyle \text{Then }all\;we\;have\;to\;do\text{ is solve the system of equations.}\)


I'll wait in the car . . .
.