# What is the equation of the locus of points at a distance of sqrt(20) units from (0,1)? What are the coordinates of the points on the line y=1/2x+1 at a distance of sqrt(20) from (0, 1)?

Jun 2, 2016

Equation: ${x}^{2} + {\left(y - 1\right)}^{2} = 20$

Coordinates of specified points: $\left(4 , 3\right)$ and $\left(- 4 , - 1\right)$

#### Explanation:

Part 1
The locus of points at a distance of $\sqrt{20}$ from $\left(0 , 1\right)$
is the circumference of a circle with radius $\sqrt{20}$ and center at $\left({x}_{c} , {y}_{c}\right) = \left(0 , 1\right)$

The general form for a circle with radius $\textcolor{g r e e n}{r}$ and center $\left(\textcolor{red}{{x}_{c}} , \textcolor{b l u e}{{y}_{c}}\right)$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{{x}_{c}}\right)}^{2} + {\left(y - \textcolor{b l u e}{{y}_{c}}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$

In this case
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {\left(y - 1\right)}^{2} = 20$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Part 2
The coordinates of the points on the line $y = \frac{1}{2} x + 1$ at a distance of $\sqrt{20}$ from $\left(0 , 1\right)$
are the intersection points of
$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{2} x + 1$ and
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {\left(y - 1\right)}^{2} = 20$

Substituting $\frac{1}{2} x + 1$ for $y$ in ${x}^{2} + {\left(y - 1\right)}^{2} = 20$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {\left(\frac{1}{2} x\right)}^{2} = 20$

$\textcolor{w h i t e}{\text{XXX}} \frac{5}{4} {x}^{2} = 20$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 16$

Either
$\textcolor{w h i t e}{\text{XXX")x=+4color(white)("XXX}} \rightarrow y = \frac{1}{2} \left(4\right) + 1 = 3$
or
$\textcolor{w h i t e}{\text{XXX")x=-4color(white)("XXX}} \rightarrow y = \frac{1}{2} \left(- 4\right) + 1 = - 1$