Finding coordinates of centre of circle

"The circle, centre (p, q) radius 25, meets the x-axis at (-7, 0) and (7, 0), where q > 0.
Find the value of p and q"

I found p = 0, since the critical points are equidistant from the origin. I then use Pythagora's theorem (Equation of the circle) to find q, but it doesn't work.

(-7 - p)^2 + (0 - q)^2 = 25

(-7 - 0)^2 + (0 - q)^2 = 25

49 + q^2 = 25
q^2 = 25 - 49
q^2 = 24

I don't know what I'm doing wrong here. By the way, the text says q should be equal to 24.

Thanks in advance :)

You have a right angled triangle with hypotenuse = 25
q^2+7^2 = 25^2
q^2 = 576
q = 24
Centre (0,24)
(You did all the hard work but didn't square the 25)

(x-p)^2 +(y-q)^2 = 25^2= 625......................(i)
x^2 -2px +p^2 +y^2 -2qy +q^2 = 625 ..........(ii)
since point(-7,0) lies on it
49 +14p +p^2 +0 +0 +q^2 = 625
p^2+q^2+14p =625-49 = 576................(iii)
Since (7,0) lies on circle
p^2 +q^2 -14 p = 576 .........................(iv)
subtract (iv) from (iii)
28 p = 0 or p = 0...............................(v)
from (iii)
0+q^2 +0 = 576
q^2 = 576
Taking sqrt
q = 24 or -24
but we have to take +24
Hence p=0 and q = 24...............Ans

Your thought process is correct. The x is 0 and the use of the pythagorean thrm use is correct. However,

x^2 + y^2 = r^2 you forgot to square the original radius.