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

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

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Answer 1 · 2016-11-23T07:10:51

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

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

Explanation:

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

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

$x^{2} + {(x - 2)}^{2} - 4 x - 2 (x - 2) = 0$ $x^2+(x-2)^2-4x-2(x-2)=0$

or $x^{2} + x^{2} - 4 x + 4 - 4 x - 2 x + 4 = 0$ $x^2+x^2-4x+4-4x-2x+4=0$

or $2 x^{2} - 10 x + 8 = 0$ $2x^2-10x+8=0$

or $x^{2} - 5 x + 4 = 0$ $x^2-5x+4=0$

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

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

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

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

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

And as the slope of radius line joining $(2, 1)$ $(2,1)$ and $(4, 2)$ $(4,2)$ is $\frac{2 - 1}{4 - 2} = \frac{1}{2}$ $(2-1)/(4-2)=1/2$ , slope of tangent at $(4, 2)$ $(4,2)$ is $- \frac{1}{\frac{1}{2}} = - 2$ $-1/(1/2)=-2$ and equation of tangent is $y - 2 = - 2 (x - 4)$ $y-2=-2(x-4)$ i.e. $y - 2 = - 2 x + 8$ $y-2=-2x+8$ or $2 x + y = 10$ $2x+y=10$ .
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]}

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