Points on a Circle

The equation of a circle in standard form is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center, and $r$ is the radius of the circle. The ordered pair $(x,y)$ represents any point on the circle. Substituting our center and radius into the equation of a circle, we have:

$(x-3{)}^2+(y-6{)}^2={13}^2$
$(x-3{)}^2+(y-6{)}^2=169$

An ordered pair $(x,y)$ is a point on the circle if it satisfies the equation of the circle. Substituting $(-2,-6)$ into the equation of the circle, we get:

$(-2-3{)}^2+(-6-6{)}^2=169$
$(-5{)}^2+(-12{)}^2=169$
$25+144=169$
$169=169$

Since the ordered pair produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is a point on the circle.
Substituting $(8,18)$ into our equation, we get:

$(8-3{)}^2+(18-6{)}^2=169$
$(5{)}^2+(12{)}^2=169$
$25+144=169$
$169=169$

Since the ordered pair $(8,18)$ produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is also a point on the circle.

Since we are given the center and a point on the circle, we can find its radius.

The equation of a circle in standard form is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center, and $r$ is the radius of the circle. The ordered pair $(x,y)$ represents any point on the circle.

Substituting our center $(-2,4)$ into the equation of the circle, we have:

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

Now entering the given point $(-2,0)$ into the equation for the circle, we have:

$(-2+2{)}^2+(0-4{)}^2=r^2$
$(-2+2{)}^2+(-4{)}^2=r^2$
$(0{)}^2+(-4{)}^2=r^2$
$16=r^2$
$\sqrt{16}=\sqrt{r^2}$
$r=4$

While $r=\pm 4$, we only use 4 since the radius has a non-negative length.

Substituting our center and radius into the equation of a circle, we have:

$(x+2{)}^2+(y-4{)}^2=4^2$
$(x+2{)}^2+(y-4{)}^2=16$

In this case, since the given point is directly below the center of the circle, we could also find the radius by counting 4 units from the center to the given point (or 4 units from the given point to the center) as shown in the figure below.

An ordered pair $(x,y)$ is a point on the circle if it satisfies the equation of the circle. Substituting $(0,4-2\sqrt{3})$ into the equation of the circle, we get:

$(0+2{)}^2+(4-2\sqrt{3}-4{)}^2=16$
$(2{)}^2+(-2\sqrt{3}{)}^2=16$
$4+4\cdot 3=16$
$4+12=16$
$16=16$

Since the ordered pair produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is a point on the circle.

Substituting $(0,4+2\sqrt{3})$ into the equation of the circle, we get:

$(0+2{)}^2+(4+2\sqrt{3}-4{)}^2=16$
$(2{)}^2+(2\sqrt{3}{)}^2=16$
$4+4\cdot 3=16$
$4+12=16$
$16=16$

Since the ordered pair $(0,4+2\sqrt{3})$ produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is also a point on the circle.

Since we are given the center and a point on the circle, we can find its radius. Then we can use the radius to find the circle’s leftmost and rightmost points.

The equation of a circle in standard form is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center, and $r$ is the radius of the circle. The ordered pair $(x,y)$ represents any point on the circle.
Substituting our center $(4,-1)$ into the equation of the circle, we have:

$(x-4{)}^2+(y-(-1){)}^2=r^2$
$(x-4{)}^2+(y+1{)}^2=r^2$>/p>

Now entering the given point $(8,2)$ into the equation for the circle, we have:

$(8-4{)}^2+(2+1{)}^2=r^2$
$(4{)}^2+(3{)}^2=r^2$
$16+9=r^2$
$25=r^2$
$\sqrt{25}=\sqrt{r^2}$
$r=5$

While $r=\pm 5$, we only use 5 since the radius has a non-negative length.

On the coordinate plane, we can use the value of $r$ to count a radius length of 5 units horizontally to the left and right (or west and east) from the center of the circle to find the furthest most left and right points, respectively.

Counting a radius length of 5 units to the west from the center produces the leftmost point of $(-1,-1)$ on the circle. Counting a radius length of 5 units to the east from the center produces rightmost point of $(9,-1)$.

Counting a radius length of 5 units from the center to the west and east always produces the leftmost and rightmost points on the circle. Below are 2 additional points that are graphed on the circle. While both points are also 5 units from the center, notice that neither are the leftmost or rightmost points.

The equation of a circle in standard form is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center of the circle, and $r$ is the radius of the circle. The ordered pair $(x,y)$ represents any point on the circle. Since the stake is at the origin of the coordinate plane, it is a point that has coordinates $(0,0)$. When the rope is fully extended, it forms the radius for the circular path your pet walks, so $r=10$. Substituting our center and radius into the equation of a circle, we have:

$(x-0{)}^2+(y-0{)}^2={10}^2$
$x^2+y^2=100$

An ordered pair (x,y) is a point on the circle if it satisfies the equation of the circle. Substituting (-8,6) into the equation of the circle, we get:

$(-8{)}^2+(6{)}^2=100$
$64+36=100$
$100=100$

Since the ordered pair produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is a point on the circle.

Substituting $(6,8)$ into our equation, we get:

$(6{)}^2+(8{)}^2=100$
$36+64=100$
$100=100$

Since the ordered pair $(6,8)$ produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is also a point on the circle.

Since the center of the hub of the bicycle is at the origin of the coordinate plane, it is a point that has coordinates $(0,0)$. We can use the origin and the given point on the circle to find its radius.

The equation of a circle in standard form is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center of the circle, and $r$ is the radius of the circle. The ordered pair $(x,y)$ represents any point on the circle. Substituting our center $(0,0)$ into the equation of the circle, we have:

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

Now entering the given point $(6,-2)$ into the equation for the circle, we have:

$(6{)}^2+(-2{)}^2=r^2$
$36+4=r^2$
$\sqrt{40}=\sqrt{r^2}$
$r=\sqrt{40}$ or $2\sqrt{10}$

While $r=\pm 2\sqrt{10}$, we only use $2\sqrt{10}$ since the radius has a non-negative length.
Substituting our center and radius into the equation, we have:

$x^2+y^2=(2\sqrt{10}{)}^2$
$x^2+y^2=4\cdot 10$
$x^2+y^2=40$

An ordered pair $(x,y)$ is a point on the circle if it satisfies the equation of the circle. Substituting $(-2,6)$ into the equation of the circle, we get:

$x^2+y^2=40$
$(-2{)}^2+(6{)}^2=40$
$4+36=40$
$40=40$

Since the ordered pair produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is a point on the circle.

Substituting $(5,\sqrt{15})$ into the equation of the circle, we get:

$(5{)}^2+(\sqrt{15}{)}^2=40$
$25+15=40$
$40=40$

Since the ordered pair $(5,\sqrt{15})$ produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is also a point on the circle.