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

1 Answer

(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