Question

A circle has a center that falls on the line `y = 2x +7` and passes through `(4 ,4 )` and `(1 ,2 )`. What is the equation of the circle?

Answer 1

`(x+1/14)^2 + (y-96/14)^2=(sqrt(4849)/14)^2`

Explanation:

The standard Cartesian form for the equation of a circle is:

`(x - h)^2 + (y-k)^2=r^2" [1]"`

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

Use equation [1] and the two points to write two equations:

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

The third equation is written by substituting the center-point to the given line:

`k = 2h+7" [4]"`

Expand equations [2] and [3]:

`16 - 8h+ h^2 + 16- 8k+k^2=r^2" [5]"`
`1 - 2h+h^2 + 4-4k+k^2=r^2" [6]"`

Subtract equation [6] from equation [5]:

`27-6h-4k=0" [7]"`

Use equation [4] to substitute into equation [7]:

`27-6h-4(2h+7)=0`

`27-6h-8h-28=0`

`-14h-1 = 0`

`h = -1/14`

Use equation [4] to find the value of k:

`k = 2(-1/14)+7`

`k = -2/14+98/14`

`k = 96/14`

Use equation [3] to find the value of r:

`(1 - -1/14)^2 + (2-96/14)^2=r^2`

`(15/14)^2+ (-68/14)^2 = r^2`

`r = sqrt(4849)/14`

Use equation [1] to write the equation of the circle:

`(x+1/14)^2 + (y-96/14)^2=(sqrt(4849)/14)^2`
graph{(27-6x-4y)(2x-3y+4)(2x-y+7)((x-4)^2+(y-4)^2-0.05) ((x-1)^2+(y-2)^2-0.05) ((x+1/14)^2+(y-96/14)^2-4849/196)=0 [-10.09, 9.91, 1.2, 11.2]}