# For the circle x^2 + y^2 = 16, what is the equation of the tangent at the point where the x-coordinate is 2.

justaguide | College Teacher | (Level 2) Distinguished Educator

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The equation of the circle is x^2 + y^2 = 16. The slope of the tangent at any point on the circle is given by `dy/dx` .

Use implicit differentiation to determine `dy/dx` . This gives:

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

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

If the x-coordinate of the point where the tangent to the circle has to be determined is 2, the y-coordinate can take 2 values `sqrt(16 - 4)` and `-sqrt(16-4)` or `2*sqrt 3` and `-2*sqrt 3`

The equation of the tangents are:

`(y - 2*sqrt 3)/(x - 2) = -2/(2*sqrt 3)`

=> `(y - 2*sqrt 3)/(x - 2) = -1/sqrt 3`

=> `sqrt 3*y - 6 = 2 - x`

=> `x + sqrt 3*y - 8 = 0`

and `(y + 2*sqrt 3)/(x - 2) = 2/(2*sqrt 3)`

=> `(y + 2*sqrt 3)/(x - 2) = 1/sqrt 3`

=> `sqrt 3*y + 6 = x - 2`

=> `x - sqrt 3*y - 8 = 0`

The equations of the required tangents are `x + sqrt 3*y - 8 = 0` and `x - sqrt 3*y - 8 = 0`

pramodpandey | College Teacher | (Level 3) Valedictorian

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Let `(x_1 ,y_1)` be a point on the circle `x^2+y62=16` .The equation of the circle at this point will be

`x x_1+yy_1=16`      (i)

In question `x_1=2 ,` so `y_1=+-sqrt(16-x^2)=+-sqrt(16-2^2)=+-sqrt(12)=+-3sqrt(3)`

``substituting values of `(x_1,y_1)=(2,+-2sqrt(3))` in (i),we have

`2x+-2sqrt(3)y=16`

`x+-sqrt(3)y=8`

Thus there are two tangents.