Question

Find the intersection point between `x^2+y^2-4x-2y=0` and the line `y=x-2` and then determine the tangent that those points?

Find the intersection point between `x^2+y^2-4x-2y=0` and the line `y=x-2` and then determine the tangent that those points?

Answer 1

Intersection points are `(1,-1)` and `(4,2)` and tangents are

`x+2y+1=0` and `2x+y=10`

Explanation:

`x^2+y^2-4x-2y=0` an be written as `(x-2)^2+(y-1)^2=5` and hence is a circle with center at `(2,1)` and radius `sqrt5`.

For finding intersection point between `x^2+y^2-4x-2y=0` and line `y=x-2`, we can put `y=x-2` in the equation `x^2+y^2-4x-2y=0`. Doing this, we get

`x^2+(x-2)^2-4x-2(x-2)=0`

or `x^2+x^2-4x+4-4x-2x+4=0`

or `2x^2-10x+8=0`

or `x^2-5x+4=0`

or `(x-1)(x-4)=0`

i.e. `x=1` or `x=4`

and for `x=1`, `y=-1` and for `x=4`, `y=2`

Hence intersection points are `(1,-1)` and `(4,2)`

As the slope of radius line joining `(2,1)` and `(1,-1)` is `(-1-1)/(1-2)=(-2)/(-1)=2`, slope of tangent at `(1,-1)` is `-1/2` and equation of tangent is `y+1=-1/2(x-1)` i.e. `2y+2=-x+1` or `x+2y+1=0`.

And as the slope of radius line joining `(2,1)` and `(4,2)` is `(2-1)/(4-2)=1/2`, slope of tangent at `(4,2)` is `-1/(1/2)=-2` and equation of tangent is `y-2=-2(x-4)` i.e. `y-2=-2x+8` or `2x+y=10`.
graph{(x^2+y^2-4x-2y)(x-y-2)(x+2y+1)(2x+y-10)=0 [-2.37, 7.63, -1.58, 3.42]}