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

Question

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

1 Answer

Impact of this question

1651 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-02T12:03:28

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

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

Explanation:

Part 1
The locus of points at a distance of $\sqrt{20}$ $sqrt(20)$ from $(0, 1)$ $(0,1)$
is the circumference of a circle with radius $\sqrt{20}$ $sqrt(20)$ and center at $(x_{c}, y_{c}) = (0, 1)$ $(x_c,y_c)=(0,1)$

The general form for a circle with radius $r$ $color(green)(r)$ and center $(x_{c}, y_{c})$ $(color(red)(x_c),color(blue)(y_c))$ is
$XXX {(x - x_{c})}^{2} + {(y - y_{c})}^{2} = r^{2}$ $color(white)("XXX")(x-color(red)(x_c))^2+(y-color(blue)(y_c))^2=color(green)(r)^2$

In this case
$XXX x^{2} + {(y - 1)}^{2} = 20$ $color(white)("XXX")x^2+(y-1)^2=20$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Part 2
The coordinates of the points on the line $y = \frac{1}{2} x + 1$ $y=1/2x+1$ at a distance of $\sqrt{20}$ $sqrt(20)$ from $(0, 1)$ $(0,1)$
are the intersection points of
$XXX y = \frac{1}{2} x + 1$ $color(white)("XXX")y=1/2x+1$ and
$XXX x^{2} + {(y - 1)}^{2} = 20$ $color(white)("XXX")x^2+(y-1)^2=20$

Substituting $\frac{1}{2} x + 1$ $1/2x+1$ for $y$ $y$ in $x^{2} + {(y - 1)}^{2} = 20$ $x^2+(y-1)^2=20$
$XXX x^{2} + {(\frac{1}{2} x)}^{2} = 20$ $color(white)("XXX")x^2+(1/2x)^2=20$

$XXX \frac{5}{4} x^{2} = 20$ $color(white)("XXX")5/4x^2=20$

$XXX x^{2} = 16$ $color(white)("XXX")x^2=16$

Either
$XXX x = + 4 XXX \to y = \frac{1}{2} (4) + 1 = 3$ $color(white)("XXX")x=+4color(white)("XXX")rarry=1/2(4)+1=3$
or
$XXX x = - 4 XXX \to y = \frac{1}{2} (- 4) + 1 = - 1$ $color(white)("XXX")x=-4color(white)("XXX")rarry=1/2(-4)+1=-1$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the equation of the locus of points at a distance of sqrt(20)√20sqrt(20) units from (0,1)(0,1)(0,1)? What are the coordinates of the points on the line y=1/2x+1y=12x+1y=1/2x+1 at a distance of sqrt(20)√20sqrt(20) from (0, 1)(0,1)(0, 1)?

1 Answer

Explanation:

Related questions

Impact of this question

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