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

Nov 23, 2016

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

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

Explanation:

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

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

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

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

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

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

or $\left(x - 1\right) \left(x - 4\right) = 0$

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

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

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

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

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