Question

Write down the equation of the tangent to the circle: `x^2+y^2-4x-6y+3=0` at the point (5,4)?

Answer 1

The equation is `y = 19 - 3x`.

Explanation:

We must take the first derivative to determine the slope of the tangent.

`2x + 2y(dy/dx) - 4 - 6(dy/dx) = 0`

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

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

`dy/dx = (2 - x)/(y -3)`

The slope of the tangent therefore at `(5, 4)` will be:

`dy/dx = (2 - 5)/(4 - 3) = -3/1 = -3`

The equation of the tangent will thus be

`y - 4 = -3(x -5)`

`y = -3x + 15 + 4`

`y = -3x + 19`

Here's an image of what is going on:

Hopefully this helps!

Answer 2

`y = -3x + 19`

Explanation:

Verify that the point `(5,4)` lies on the circle:

`5^2+4^2-4(5)-6(4)+3=0`

`0=0 larr` verified.

To find the slope of the tangent line, we must implicitly differentiate the equation of the circle:

`(d(x^2))/dx + (d(y^2))/dx -4(d(x))/dx -6dy/dx + (d(3))/dx = 0`

Perform the differentiations:

`2x + 2ydy/dx -4 -6dy/dx = 0`

Move all of the terms that do NOT contain `dy/dx` to the right:

`2ydy/dx -6dy/dx = 4-2x`

Factor out `dy/dx`:

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

Divide both sides by `(2y -6)`

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

The slope, m, of the tangent line, is the above evaluated at `(5,4)`:

`m = (4-2(5))/(2(4) -6)`

`m = -3`

The point-slope form of the equation of a line is:

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

Substitute `m = -3`, `x_1 = 5`, and `y_1 = 4`:

`y = -3(x - 5)+4`

Simplify:

`y = -3x + 19`