What is the value of a if the area of a circle formed by the points (7, 0), (11, 0) and (0, a) is 36 squared cm.

Since  the points (7,0) and(11,0) are the points on the
circle,

Let (h,k) be the center . Then  the centre (h,k) is
equidistant from (7,0) and (11,0).

So,  (h-7)^2 +(k-0)^2 =
(h-11)^2+k(0)^2 , k^2 gets cancelled on both sides.

h^2-14h+49 =
h^2-22h+121,  h^2 gets cancelled on both sides.

-14h+49 =
-22h+121

22h-14h = 121-49 = 72.

8h =
72, or h = 72/8 = 9.

Also are of the
circle is 36 sq cm.

Therefore pir^2 = 36. So r^2 =
36/pi.

Now since  (7,0) is on the
circle,

(7-h)^2 +(0-k)^2 =
r^2.

Therefore  k^2 = r^2-(7-h)^2 = r^2 -(7-9)^2 =
r^2-4.

So k^2 = (36/pi - 4).

Therefore
k = sqrt(3/pi-4) , Or k = -sqrt(36/pi -4).

To find the value of a if
(0, a) is a point on the circle.

Now the centre of the circle is
(h,k) = (9 , sqrt(36/pi -4) ) or  (9 , -sqrt(36/pi -4) ).

(0,a) is
point on the circle with centre ( 9, sqrt(36/pi -4) )

 (0-9)^2 +
(a-sqrt(36/pi - 4) ^2  = 36/pi

81 + a^2 -2asqrt(36/-4) + 36/pi -4 =
36/pi

a^2 -2a sqrt(36/pi-4) -4 = 0

 a1
= {2+ sqrt { 36/pi-4 +16)}/2 = {1+sqrt(36/pi +12)}

 a2 =
{1-sqrt(36/pi +12).