A circle has a center that falls on the line #y = 5/8x +6 # and passes through # ( 1 ,5 )# and #(2 ,9 )#. What is the equation of the circle?

Question

A circle has a center that falls on the line #y = 5/8x +6 # and passes through #      (    1  ,5  )# and #(2   ,9  )#.  What is the equation of the circle?

Sadie Adams · Accepted Answer

#(x-11/7)^2+(y-391/56)^2=\frac{13345}{3136}#

Let #(x_1, y_1)# be the center & #r# be the radius of circle.

The center #(x_1, y_1)# lies on the line: #y=5/8x+6# hence coordinates of center will satisfy the equation of line as follows

#y_1=5/8x_1+6\ ...........(1)#

Now, using distance formula the distance between the center #(x_1, y_1)# & the point #(1, 5)# will be equal to radius #r# as follows

#\sqrt{(x_1-1)^2+(y_1-5)^2}=r#

#(x_1-1)^2+(y-5)^2=r^2\ .......(2)#

Similarly, using distance formula the distance between the center #(x_1, y_1)# & the point #(2, 9)# will be equal to radius #r# as follows

#\sqrt{(x_1-2)^2+(y_1-9)^2}=r#

#(x_1-2)^2+(y-9)^2=r^2\ .......(3)#

As you can see, subtracting (3) from (2)

#(x_1-1)^2+(y-5)^2-(x_1-2)^2-(y-9)^2=r^2-r^2#

#2x_1+8y_1=59#

Substituting value of #y_1# from (1) in the above equation, we get

#2x_1+8(5/8x_1+6)=59#

#2x_1+5x_1+48=59#

#7x_1=11#

#x_1=11/7#

setting value of #x_1# in (1) as follows

#y_1=5/8(11/7)+6#

#y_1=391/56#

setting values of #x_1# & #y_1# in (2), we get

#(11/7-1)^2+(391/56-5)^2=r^2#

#r^2=\frac{13345}{3136}#

Consequently, the circle equation is

#(x-11/7)^2+(y-391/56)^2=\frac{13345}{3136}#

Aurora Ayala · Answer

undefined The equation of the circle is $ (x - 4)^2 + (y - 3)^2 = 5 $.