A circle in the standard (x,y) coordinate plane intersect x-axis at (-7,0) and (1,0). The radius of the circle is 5 coordinates units.?

The general eq'n for a circle is
x^2 + y^2 + 2fx + 2gy = r^2
Hence
(x - -7)^2 + (y - 0)^2 = 5^2
(x + 7)^2 + ( y - 0)^2 = 5^2
x^2 + 14x + 49 + y^2 = 5^2
&
( x - 1)^2 + y^2 = 5^2
x^2 - 2x + 1 + y^2 = 5^2
Subtract the two circular eq'ns
16x + 48 = 0
16x = -48
x = -3
(-3)^2 - 2(-3) + 1 + y^2 = 25
9 + 6 +1 + y^2 =25
16 + y^2 = 25
y^2 = 9 =
y = +/- sqrt(9)
y = +/- 3

So the circle centre could be either ( -3, 3) or ( -3, -3)

(-3,0) is midway between x-intercepts
Length of the apothem is 3 units.
center (-3,3) or (-3,-3)

Let the circle be (x-h)^2+(y-k)^2=25, then
(-7,0) on the circle=>(-7-h)^2+k^2=25------(1)
(1-h)^2+k2=25------(2)
(1)-(2)=>
(7+h)^2-(1-h)^2=0=>
h=-3------(3)
from (2) & (3), get
k=+/-3
=> the center of the circle=(-3, 3) & (-3, -3).
Ans. A & C

The distance from (1,0) to (-3,0) is not 5, so (-3,0) cannot be the center. Each of the other two works OK (is 5 units away from both of the points (1,0) and (-7,0)).