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is a point on the
x-axis. If
(-3,5) was the center of the circle with radius 13 units, find all
t possible coordinates for
P. (Try this problem algebraically without formally graphing)

is a point on the $x$ -axis. If $(-3,5)$ was the center of the circle with radius $13$ units, find all $t$ possible coordinates for $P$ . (Try this problem algebraically without formally graphing)

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Write equations of circles in standard form using properties

Full solution

Q. is a point on the

x

-axis. If

(-3,5)

was the center of the circle with radius

13

units, find all

t

possible coordinates for

P

. (Try this problem algebraically without formally graphing)

Circle Equation: The equation of a circle with center $(h,k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$ . Here, $h = -3$ , $k = 5$ , and $r = 13$ .
Substitute Values: Substitute the values into the equation: $(x - (-3))^2 + (y - 5)^2 = 13^2$ .
Simplify Equation: Simplify the equation: $(x + 3)^2 + (y - 5)^2 = 169$ .
Substitute $y=0$ : Since $P$ is on the $x$ -axis, $y = 0$ . Substitute $y = 0$ into the equation: $(x + 3)^2 + (0 - 5)^2 = 169$ .
Simplify Equation: Simplify the equation: $(x + 3)^2 + 25 = 169$ .
Subtract $25$ : Subtract $25$ from both sides: $(x + 3)^2 = 144$ .
Take Square Root: Take the square root of both sides: $x + 3 = \pm 12$ .
Solve for x: Solve for x: $x = -3 \pm 12$ .
Find x Values: Find the two possible values for x: $x = -3 + 12$ and $x = -3 - 12$ .
Calculate x: Calculate the values: $x = 9$ and $x = -15$ .

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