What's the tangent's gradient?

Relevant wiki: Differentiation Rules

Differentiation can be applied here. The differential function of a curve and the tangent to the curve at a point are related.

For example: lets take the parabola y = x $^{2}$ +5x, differentiating the function gives $\frac{dy}{dx}$ = 2x + 5.

A tangent passing through the point (1,6) on the parabola will have a gradient $m$ = $\frac{dy}{dx}$ (1). thus $m$ = $\frac{dy}{dx}$ (1) = 2(1) +5 = 2+5 =7

In conclusion, the gradient of a tangent passing through (x,y) in a curve is equal to $\frac{dy}{dx}$ (x,y) where $\frac{dy}{dx}$ is the differential function of the curve.

In the given equation: x $^{2}$ + y $^2$ - 8x + 2y =1. differentiating the function will result to: 2x +2y $\frac{dy}{dx}$ - 8 + 2 $\frac{dy}{dx}$ = 0.

Rearranging to make $\frac{dy}{dx}$ subject; 2y $\frac{dy}{dx}$ + 2 $\frac{dy}{dx}$ = 8 - 2x
(2y + 2) $\frac{dy}{dx}$ = 8 - 2x
$\frac{dy}{dx}$ = $\frac{8 - 2x}{2y + 2}$

$\frac{dy}{dx}$ = $\frac{4 - x}{y + 1}$ .

next we substitute the point (1,2) and get $m$ = (4 - 1)/( 2 + 1) = 3/3 = 1 The answer is 1

Alternatively , The equation of the circle can be simplified to (x - 4) $^{2}$ + (y + 1) $^{2}$ = 18

This shows that the circle has center at (4 , -1) and radius $sqrt{18}$

Since the tangent of a circle is perpendicular to the radius at a point on the circle, we simply find the gradient of the radius( from the center (4 , -1) to the given point (1,2) i.e $\frac{2 - (-1)}{1 - 4}$ = -1

Finally, from coordinate geometry, product of perpendicular lines equal -1. Therefore, the gradient of the tangent (which is perpendicular to the radius) is -1/-1 which is 1

3rd Alternative Find the coordinates for the centre of the circle, Then, find the slope to the given point in the question. the tangent is perpendicular to this slope, so find the value of the slope perpendicular to the one found earlier on. that's the answer