Question

How do you write the equation of the line that is tangent to the circle `(x + 6)^2 + (y + 4)^2 = 25` at the point (-9, -8)?

Answer 1

We will have to differentiate the function through implicit differentiation to find this one.

Let's first distribute:

`x^2 + 12x + 36 + y^2 + 8y + 16 = 25`

`x^2 + y^2 + 12x + 8y + 52 - 25 = 0`

`x^2 + y^2 + 12x + 8y + 27 = 0`

`x^2 + y^2 + 12x + 8y = -27`

We start by differentiating both sides:

`d/dx(x^2 + y^2 + 12x + 8y) = d/dx(-27)`

`2x + 2y(dy/dx) + 12 + 8(dy/dx) = 0`

`2y(dy/dx) + 8(dy/dx) = -12 - 2x`

`dy/dx(2y + 8) = -12 - 2x`

`dy/dx = (-12 - 2x)/(2y + 8)`

`dy/dx = -(2(6 + x))/(2(y + 4))`

`dy/dx = -(x + 6)/(y + 4)`

The slope of the tangent is given by inserting our given point, `(x, y)`, into the derivative.

Hence, `m_"tangent" = -(-9 + 6)/(-8 + 4) = -(-3/-4) = - 3/4`

The equation of the tangent can now be found by using point-slope form, since this will be a straight line.

`y - y_1 = m(x - x_1)`

`y - (-8) = -3/4(x - (-9))`

`y + 8 = -3/4x - 27/4`

`y = -3/4x - 27/4 - 8`

`y = -3/4x - 59/4`

So, the equation of the tangent to `y = (x + 6)^2 + (y + 4)^2 = 25` at the point `(-9, -8)` is `y = -3/4x - 59/4`.

Hopefully this helps!