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

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#