# For the circle x^2 + y^2 = 49 what is the slope of the tangent at the point where x = 1.

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### 2 Answers

The slope of the tangent to the circle given by the equation x^2 + y^2 = 49 at the point where x = 1 has to be delivered.

Determine the derivative `dy/dx` for the given equation. Using implicit differentiation gives:

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

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

The value of the derivative `dy/dx` at any point on the circle is the slope of the tangent at that point.

For x = 1, there are two possible values of y.

1 + y^2 = 49

=> y = `+-4*sqrt 3`

This gives two values of `dy/dx` as there are two points on the given circle where x = 1.

`dy/dx = -1/(4*sqrt 3)` and `dy/dx = 1/(4*sqrt 3)`

**The slope of the tangent to the circle x^2 + y^2 = 49 at one of the points where x = 1 is `-1/(4*sqrt 3)` and at the other point it is `1/(4*sqrt 3)` **

Given the circle x^2 + y^2 = 49 we need to find the slope of the tangent at the point where x = 1.

First use implicit differentiation.

`2x + 2ydy/dx = 0`

`dy/dx = -x/y`

When x = 1, y = `sqrt(48)`

`Plug these two values in the derivative to get the slope. `

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