# What is the equation of the tangent to the curve x^2 + y^2 = 36 that passes through the point (6, 6)

*print*Print*list*Cite

### 1 Answer

The equation of the curve that is given is x^2 + y^2 = 36. This is the equation of a curve that has the origin as its center and a radius of 6.

If a tangent is drawn on the circle at any point, the slope of the tangent is the value of the derivative `dy/dx` at that point.

Using implicit differentiation, `2x + 2y*(dy/dx) = 0`

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

If a line tangent to the circle at a point (x, y) passes through the point (6, 6) it has a slope given by `(6 - y)/(6 - x)` . This is also equal to `-x/y`

`(6 - y)/(6 - x) = -x/y`

=> `6y - y^2 = -6x + x^2`

=> `x^2 + y^2 = 6x + 6y`

Substituting `x^2 + y^2 = 6x + 6y` in the equation of the circle gives 6x + 6y = 6

=> x + y = 6

As (x, y) lies on the circle x^2 + y^2 = 36 either x = 0 and y = 6 or x = 6 and y = 0

This gives the points at which tangent lines from (6, 6) touch the circle as (0, 6) and (6, 0)

The equations of the tangents are `(y - 6)/(x - 6) = 0` which is a horizontal line y = 6 and `(y - 6)/(x - 6) = 6/0` which is a vertical line x = 6.

**The equations of the tangents to the curve x^2 + y^2 = 36 that pass through (6, 6) are x = 6 and y = 6**