# 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

*print*Print*list*Cite

### 1 Answer

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**