Question

How do you find the equation of the circle passing through (7,5) and (3,7), and with center on x-3y+3=0?

Answer 1

`(x-3)^2+(y-2)^2=25`

Explanation:

Let `(x_1, y_1)` be the center & `r` be the radius of unknown circle then as per given conditions, the distance of each of points `(7, 5)` & `(3, 7)` from center `(x_1, y_1)` will be equal to radius `r` as follows

`(x_1-7)^2+(y_1-5)^2=r^2\ .......(1)`

`(x_1-3)^2+(y_1-7)^2=r^2\ .......(2)`

Subtracting (1) from (2), we get

`2x_1-y_1-4=0\ ..............(3)`

Since, the center `(x_1, y_1)` of circle lies on the straight line: `x-3y+3=0` hence it will satisfy the equation f straight line as follows

`x_1-3y_1+3=0\ ........(4)`

solving (3) & (4), we get `x_1=3, y_1=2`

substituting the values of `x_1=3, y_1=2` in (1), we get `r^2=25`

hence, the equation of circle is

`(x-x_1)^2+(y-y_1)^2=r^2`

`(x-3)^2+(y-2)^2=25`