Question

What is the equation of the circle that passes through points `(0,1)`, `(1,4)` and `(5,2)`?

Answer 1

`(x-17/7)^2+(y-13/7)^2=((5sqrt13)/7)^2`

Explanation:

The Cartesian equation for a circle is of the form:

`(x-h)^2+(y-k)^2=r^2`

where (x,y) is any point on the circle, `(h,k)` is the center, and `r` is the radius.

We can use this form and the 3 points `(0,1)`, `(1,4)` and `(5,2)` to write three equations:

`(0-h)^2+(1-k)^2=r^2" [1]"`
`(1-h)^2+(4-k)^2=r^2" [2]"`
`(5-h)^2+(2-k)^2=r^2" [3]"`

Expand the squares:

`h^2+1-2k+k^2=r^2" [1.1]"`
`1-2h+h^2+16-8k+k^2=r^2" [2.1]"`
`25-10h+h^2+4-4k+k^2=r^2" [3.1]"`

Subtract equation [2.1] from equation [1.1]:

`2h+6k-16=0" [4]"`

Subtract equation [3.1] from equation [1.1]:

#10h+2k-28=0" [5]"

Multiply equation [5] by -3 and add to equation [4]:

`-28h + 68 = 0`

`h = 17/7`

Substitute the value for h into equation [4] to solve for k:

`2(17/7)+6k-16=0`

`6k=16-34/7`

`6k = 78/7`

`k = 13/7`

Substitute the values for h and k into equation [1] to solve for `r`:

`(17/7)^2+(1-13/7)^2=r^2`

`(17/7)^2+(-6/7)^2=r^2`

`289/49+36/49=r^2`

`r^2 = 325/49`

`r = (5sqrt13)/7`

Substitute the values for h, k and r into the Cartesian form:

`(x-17/7)^2+(y-13/7)^2=((5sqrt13)/7)^2`

Answer 2

`x^2+y^2-34/7x-26/7y+19/7 = 0`

Explanation:

I normally solve these kind of problems geometrically, but let's just do it using systems of equations instead.

Suppose the equation of our circle is:

`x^2+y^2+ax+by+c = 0`

Since this is satisfied by each of the given points we find:

`0 = color(blue)(0)^2+color(blue)(1)^2+color(blue)(0)a+color(blue)(1)b+c = 1+b+c`

`0 = color(blue)(1)^2+color(blue)(4)^2+color(blue)(1)a+color(blue)(4)b+c = 17+a+4b+c`

`0 = color(blue)(5)^2+color(blue)(2)^2+color(blue)(5)a+color(blue)(2)b+c = 29+5a+2b+c`

Subtracting the first equation from the second and third equations, we get:

`0 = 16+a+3b`

`0 = 28+5a+b`

Multiplying the second of these equations by `3` we get:

`0 = 84+15a+3b`

Then subtracting the first, we get:

`0 = 68+14a`

Hence:

`a = -68/14 = -34/7`

Then:

`b = -28-5a = -28-5(-34/7) = -26/7`

and:

`c = -b-1 = 19/7`

So our circle has equation:

`x^2+y^2-34/7x-26/7y+19/7 = 0`

graph{x^2+y^2-34/7x-26/7y+19/7 = 0 [-8.92, 11.08, -2.48, 7.52]}