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?

1 Answer
Nov 23, 2016

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]}