Question

What is the slope of the curve `x^2+y^2-12x+4y-5=0` at `(0,1)`?

Answer 1

The slope is `=2`

Explanation:

Our function is

`x^2+y^2-12x+4y-5=0`

Differentiating with respect to `x`

`2x+2ydy/dx-12+4dy/dx=0`

`dy/dx(2y+4)=12-2x`

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

At the point `(0,1)`

`dy/dx=6/3=2`

graph{(x^2+y^2-12x+4y-5)(y-1-3x)=0 [-18.35, 22.22, -9.55, 10.76]}

Answer 2

Slope of curve is `2`.

Explanation:

As theree is no term containing `xy` andcoefficients of `x^2` and `y^2` are equal, `x^2+y^2-12x+4y-5=0` is the equation of circle and as `(0,1)` satisfiees the equation as

`0^2+1^2-12xx0+4xx1-5=0+1-0+4-5=0`, `(0,1)` lies on the circle.

Now the slope of the curve is te value of `(dy)/(dx)`, which is given by

`2x+2y(dy)/(dx)-12+4(dy)/(dx)=0`

or `(dy)/(dx)=(12-2x)/(2y+4)`

and at `(0,1)`. `(dy)/(dx)=12/6=2`

Further, slope of curve is the same as slope of tangent at that point, which would be `y-1=2x` i.e. `y=2x+1`.

graph{(x^2+y^2-12x+4y-5)(y-2x-1)=0 [-5.085, 14.915, -4.64, 5.36]}