What is b if the two curves have a common tangent : x^2 + b^2 – 2bx + y^2 – 4y - 45 = 0 , x^2 + y^2 – 2x + 20y + 65 = 0

We have the two curves: x^2 + b^2 – 2bx + y^2 – 4y - 45 =
0 , x^2 + y^2 – 2x + 20y + 65 = 0

We see that the two
curves are equations of circles. If they have a common tangent the distance between
their centers should be equal to the sum of their
radius.

• x^2 + b^2 – 2bx + y^2 – 4y - 45 =
  0

=> x^2 + b^2 – 2bx + y^2 + 4 – 4y =
49

=> (x – b)^2 + ( y – 2)^2 =
49

=> (x – b)^2 + ( y – 2)^2 =
7^2

• x^2 + y^2 – 2x + 20y + 65 =
  0

=> x^2 + 1 – 2x + y^2 + 100 + 20y =
36

=> (x - 1)^2 +( y + 10)^2 =
36

=> (x - 1)^2 +( y + 10)^2 =
6^2

The centers of the circles are ( b , 2) and ( 1 , -10).
Their radius is 7 and 6 respectively.

Equating the distance
between the centers and the sum of the radius we get:

sqrt
[ ( b – 1)^2 + ( 2 + 10)^2] = 13

sqrt [ (b – 1)^2 + 12^2 ]
= 13

=> (b – 1)^2 + 12^2 =
13^2

Instead of solving for b using quadratic equations we
can use the fact that 5 , 12 and 13 form a Pythagorean
Triplet.

So (b – 1) =
5

=> b = 5 +
1

=> b =
6

The required value of b is
6