A circle has a center that falls on the line #y = 2/3x +7 # and passes through #(5 ,7 )# and #(1 ,2 )#. What is the equation of the circle?

Question

A circle has a center that falls on the line #y = 2/3x +7 # and passes through #(5  ,7  )# and #(1   ,2  )#.  What is the equation of the circle?

Sadie Adams · Accepted Answer

see below

the centre of the circle is on the intersection point between the assigned line and the line axis of the segment that joint the two points straight for two points: #(y-y_a)/(y_b-y_a) = (x-x_a)/(x_b-x_a)# #(y-7)/(2-7) = (x-5)/(1-5)# #(y-7)/-5 = (x-5)/-4# #(y-7)/-5 = (x-5)/-4# #-4 (y-7) = -5 (x-5))# #-4 y +28 = -5 x+ 25# #-4 y = -5 x-3# #y= 5/4 x + 3/4# middle point M #X_m =( X_b+X_a) /2=(5+1)/2=3# #Y_m =( Y_b+Y_a) /2=(7+2)/2=9/2# perpendicular line (m =-1/m') passing through M #y-y_M= m (x-x_m)# #y-9/2= -4/5(x-3)# #y= -4/5 x +12/5+9/2# #y= -4/5 x +59/10# making the system and solving with comparison #2/3 x +7= -4/5 x +69/10# #22/15 x =-1/10# #X_C=-3/22# #Y_C=-1/11+7=76/11# #R=sqrt((X_c-X_a)^2 + (Y_c-Y_a)^2)=sqrt((-3/22-1)^2 + (76/11-2)^2)=#... I'm running out of time. equation of circle #(x-x_c)^2 + (y-y_c)^2 = R^2#

Sadie Allen · Answer

#(x+3/44)^2+(y-153/22)^2=49733/1936#.

Let, #C(h,k)# and #r# be the centre and radius of the reqd. circle #S.# Since, #C# is on the line #y=2/3x+7," we have, "k=2/3h+7#. #:. C(h,k)=C(h,2/3h+7)#. The points #P(5,7) and Q(1,2) in S#. # :. CP^2=r^2=CQ^2#. #:.(h-5)^2+(2/3h+7-7)^2=(h-1)^2+(2/3h+7-2)^2#. #:.(h^2-10h+25)+4/9h^2=h^2-2h+1+4/9h^2+20/3h+25#. #:.-10h+2h-20/3h=1#. #:. h=-3/44#. #:. k=2/3h+7=2/3(-3/44)+7=153/22#. Finally, #r^2=(h-5)^2+(2/3h)^2=(-3/44-5)^2+(-1/22)^2#. #;. r^2=49733/1936#. #:. S : (x-h)^2+(y-k)^2=r^2, i.e., # #(x+3/44)^2+(y-153/22)^2=49733/1936#.

Allison Allen · Answer

undefined To find the equation of the circle, first, determine the center of the circle by finding the intersection point of the line $ y = \frac{2}{3}x + 7 $ and the perpendicular bisector of the line segment joining the given points (5, 7) and (1, 2). Then, calculate the radius of the circle using the distance formula between the center and one of the given points. Finally, use the center and radius to form the equation of the circle in the standard form: $ (x - h)^2 + (y - k)^2 = r^2 $, where (h, k) is the center and r is the radius.