Question

How do you write an equation of the line tangent to `x^2+y^2=169` at the point (5,12)?

Answer 1

`y=1/12(-5x+169)`

Explanation:

Here we have the equation of a circle: `x^2+y^2 = 13^2`

To determine the slope of a tangent to the circle at any point we need to use implicit differentiation.

`x^2+y^2 = 13^2`

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

`dy/dx = -x/y`

At the point `(5, 12)`, `dy/dx= -5/12`

So the tangent has a slope of `-5/12` and passes through the point `(5, 12)`

The equation of a line of slope m, passing through the point `(x_1, y_1)` is: `(y-y_1) = m(x-x_1)`

The tangent would therefore have the equation:

`(y-12) = -5/12(x-5)`

`12y-144 =-5x+25`

`12y=-5x+169`

`y=1/12(-5x+169)`

Answer 2

`5x+12y = 169`

Explanation:

Differentiate implicitly:

`d/dx (x^2+y^2) = 0`

`2x+2ydy/dx = 0`

`dy/dx = -x/y`

So, for `x=5`, `y=12`:

`y'(5) = [dy/dx]_(5,12) = -5/12`

The equation of the tangent line is:

`y= y_0+y'(x_0)(x-x_0) = 12-5/12(x-5)`

`y=-5/12x +(144+25)/12`

`12y+5x = 169`

Answer 3

`12y+5x=169`

Explanation:

Another way without using Calculus.

The centre of the circle is `(0,0)`

so the gradient of the Normal at the point of contact is

`m=(12-0)/(5-0)=12/5`

( the radius is perpendicular to the tangent.)

for perpendicular lines the product of their gradients is `-1`

`m_txxm_n=-1`

`:.m_txx12/5=-1`

`=>m_t=-5/12`

using

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

`y-12=-5/12(x-5)`

`12y-144=-5x+25`

`12y+5x=169`