Question

A circle has a center that falls on the line `y = 1/3x +7` and passes through `( 3 ,1 )` and `(7 ,9 )`. What is the equation of the circle?

Answer 1

The equation of the circle is:

`(x-(-39/5))^2+(y-22/5)^2=(2/5sqrt281)^2`

Explanation:

Use the standard Cartesian form of the equation of a circle

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

and the two points to write two equations:

`(3-h)^2+(-6-k)^2=r^2" [1]"`
`(7-h)^2+(2-k)^2=r^2" [2]"`

Expand the squares:

`9-6h+ h^2+36+12k+k^2=r^2" [1.1]"`
`49-14h+h^2+4-4k+k^2=r^2" [2.1]"`

Subtract equation [2.1] from equation [1.1]

`-40+8h+32+16k = 0`

Combine like terms:

`8h+16k = 8`

Divide both sides by 8:

`h+2k = 1" [3]"`

Evaluate the equation `y = 1/3x +7` at the center point `(h,k)`:

`k = 1/3h +7" [4]"`

This allows us to substitute `1/3h + 7` into equation [3]:

`h+2(1/3h + 7) = 1`

`h = -39/5`

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

`k = 1/3((-39)/5) +7`

`k= 22/5`

Substitute the values of h and k into equation [1]:

`(3-(-39/5))^2+(-6-22/5)^2=r^2`

`r = 2/5sqrt281`

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

`(x-(-39/5))^2+(y-22/5)^2=(2/5sqrt281)^2`