No. $\newline$ Date $\newline$ Exercises $\newline$ $1$ . Is the point $(0,4)$ inside or autside the circle of radius \& with centre at $(-3,1)$ \& $\newline$ $2$ . Determine y so that $(0, y)$ shall be on the circle of radius $4$ with centre at $(-3,1)$ ? $\newline$ $3$ . If ane end of a line whose length is $13$ arits is the point $(4,8)$ and the ordinate of the other end is $3$ . What is its abscissa?

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Algebra 2

Find Coordinate on Unit Circle

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Q. No.

\newline

Date

\newline

Exercises

\newline

1

. Is the point

(0,4)

inside or autside the circle of radius \& with centre at

(-3,1)

\newline

2

. Determine y so that

(0, y)

shall be on the circle of radius

4

with centre at

(-3,1)

\newline

3

. If ane end of a line whose length is

13

arits is the point

(4,8)

and the ordinate of the other end is

3

. What is its abscissa?

Check Point Location: To check if $(0,4)$ is inside or outside the circle, use the distance formula to find the distance from the point to the center of the circle and compare it to the radius. $\newline$ Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $\newline$ Distance = $\sqrt{(0 - (-3))^2 + (4 - 1)^2}$ $\newline$ Distance = $\sqrt{(3)^2 + (3)^2}$ $\newline$ Distance = $\sqrt{9 + 9}$ $\newline$ Distance = $\sqrt{18}$ $\newline$ Distance = $3\sqrt{2}$ , which is approximately $4.24$ . $\newline$ Since $4.24$ is greater than the radius $4$ , the point $(0,4)$ is outside the circle.
Find y on Circle: To find $y$ so that $(0,y)$ is on the circle, use the equation of the circle $(x - h)^2 + (y - k)^2 = r^2$ , where $(h,k)$ is the center and $r$ is the radius. $\newline$ $(x + 3)^2 + (y - 1)^2 = 4^2$ $\newline$ Substitute $x = 0$ into the equation. $\newline$ $(0 + 3)^2 + (y - 1)^2 = 16$ $\newline$ $9 + (y - 1)^2 = 16$ $\newline$ $(y - 1)^2 = 16 - 9$ $\newline$ $(0,y)$ $0$ $\newline$ $(0,y)$ $1$ $\newline$ $(0,y)$ $2$ $\newline$ So the two possible values for $y$ are $(0,y)$ $4$ and $(0,y)$ $5$ .
Find Abscissa: To find the abscissa of the other end of the line segment, use the distance formula where one end is $(4,8)$ and the other end is $(x,3)$ , and the distance is $13$ .
$\text{Distance} = \sqrt{(x - 4)^2 + (3 - 8)^2}$
$13 = \sqrt{(x - 4)^2 + (-5)^2}$
$13 = \sqrt{(x - 4)^2 + 25}$
$169 = (x - 4)^2 + 25$
$144 = (x - 4)^2$
$x - 4 = \pm12$
$x = 4 \pm 12$
So the two possible values for $x$ are $16$ and $-8$ .