How do you find the equation of a circle whose centre lies on the line 2x+y-1=0 and which passes through the point A(-2,0) AND B(5,1)?

The standard form for the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2#

where #(x,y)# is any point on the circle, #(h,k)# is the center point and #r# is the radius.

Use the standard form to write two equations using points A and B:

#(-2 - h)^2 + (0 - k)^2 = r^2#

#(5 - h)^2 + (1 - k)^2 = r^2#

Because #r^2 = r^2#, we can set the left sides equal:

#(-2 - h)^2 + (0 - k)^2 = (5 - h)^2 + (1 - k)^2#

Expand the squares using the pattern #(a - b)^2 = a^2 - 2ab + b^2#

#4 + 4h + h^2 + k^2 = 25 - 10h + h^2 + 1 - 2k + k^2#

Combine like terms (noting that the squares cancel):

#4 + 4h = 25 - 10h + 1 - 2k#

Move the k term the left and all other terms to the right:

#2k = -14h + 22#

Divide by 2

#k = -7h + 11# [1]

Evaluate the given line at the center point:

#2h + k - 1 = 0#

Write in slope-intercept form

#k = -2h + 1 # [2]

Subtract equation [2] from equation [1]:

#k - k = -7h + 2h + 11 - 1#

#0 = -5h + 10#

#h = 2#

Substitute 2 for h in equation [2]

#k = -2(2) + 1 #

#k = -3#

Substitute the center #(2,-3)# into the equation of a circle using point A and solve for the value of r:

#(-2 - 2)^2 + (0 - -3)^2 = r^2#

#(-4)^2 + 3^2 = r^2#

#r^2 = 25#

#r = 5#

Substitute the center #(2, -3)# and #r = 5 into the general equation of a circle, to obtain the specific equation for this circle:

#(x - 2)^2 + (y - -3)^2 = 5^2#