# Using calculus determine the radius of the circle x^2 + 4x + y^2 + 2y - 8 = 0

### 1 Answer | Add Yours

The equation x^2 + 4x + y^2 + 2y - 8 = 0 is that of a circle. The radius of the circle has to be determined using calculus.

The distance between the point where the value of y is maximum and that at which the value of y is minimum is the diameter of the circle. To find these points first find `dy/dx`

x^2 + 4x + y^2 + 2y - 8 = 0

Use implicit differentiation

=> `2x + 4 + 2y*(dy/dx) + 2*(dy/dx) = 0`

=> `(dy/dx)(2y + 2) = -(2x + 4)`

=> `(dy/dx) = -(x + 2)/(y + 1)`

At the extreme points `dy/dx = 0`

=> x = -2

As x is same for both the points, determine the values of y

=> y^2 + 2y - 8 + 4 - 8 = 0

=> y^2 + 2y - 12 = 0

=> y1 = 1 - sqrt 13 and y2 = -1 - sqrt 13

y1 - y2 = 2*sqrt 13

The diameter of the circle is 2*sqrt 13.

**The radius of the circle is **`sqrt 13`