# What is the equation of the tangent to the curve x^2 + y^2 = 2 that passes though the point (4, 0).

### 1 Answer | Add Yours

The equation of the curve that is given is x^2 + y^2 = 2. This is the equation of a circle with center (0,0) and radius `sqrt 2` .

The slope of a tangent at any point (x, y) on the circle is the value of `dy/dx` at that point. Using implicit differentiation, `dy/dx` for the curve is given by:

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

=> `dy/dx = -x/y`

If a line passes through the point (4, 0) and a point (x, y) lying on the circle:

`(y - 0)/(x - 4) = -x/y`

=> `y^2 = 4x - x^2`

Substitute `y^2 = 4x - x^2` in the equation of the circle. This gives `x^2 + 4x - x^2 = 0`

=> x = 0

Substituting x = 0 in the equation of the circle gives y^2 = 2

=> y = `sqrt 2` and y = `-sqrt 2`

The equation of the required tangents is that of the line passing through `(4, 0)` and `(0, sqrt 2)` and the line passing through `(4, 0)` and `(0, -sqrt 2)`

This gives the equation of one tangent as: `(y - 0)/(x - 4) = (sqrt 2/-4)`

=> `-4y = sqrt 2*x - 4*sqrt 2`

=> `x + 2*sqrt 2y - 4 = 0`

and the other tangent as:

`(y - 0)/(x - 4) = (-sqrt 2/-4)`

=> `-4y = -sqrt 2*x + 4*sqrt 2`

=> `x - 2*sqrt 2y - 4 = 0`

**The equation of the tangents to the circle x^2 + y^2 = 2 that pass through (4, 0) is `x + 2*sqrt 2y - 4 = 0` and **`x + 2*sqrt 2y - 4 = 0`