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(x-7)^(2)+(y+5)^(2)=16
The given equation represents a circle in the
xy-plane. If
(x,-5) is a point on the circle, what is a possible value of
x ?

◻

$(x-7)^{2}+(y+5)^{2}=16$ $\newline$ The given equation represents a circle in the $xy$ -plane. If $(x,-5)$ is a point on the circle, what is a possible value of $x$ ? $\newline$ ◻

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Write a quadratic function from its x-intercepts and another point

Full solution

Q.

(x-7)^{2}+(y+5)^{2}=16

\newline

The given equation represents a circle in the

xy

-plane. If

(x,-5)

is a point on the circle, what is a possible value of

x

?

\newline

◻

Given Circle Equation: We are given the equation of a circle: x\(-7)^ $2$ + (y+ $5$ )^ $2$ = $16$ \. We are also given a point on the circle where the y-coordinate is (-5\)\. We need to find the corresponding x-coordinate.
Substitute $y = -5$ : Substitute $y = -5$ into the equation of the circle to find the possible values of $x$ . $(x-7)^2 + (-5+5)^2 = 16$ $(x-7)^2 + 0 = 16$ $(x-7)^2 = 16$
Take Square Root: Take the square root of both sides of the equation to solve for $x$ . $\sqrt{(x-7)^2} = \pm\sqrt{16}$ $x - 7 = \pm4$
Solve for x: Solve for x by adding $7$ to both sides of the equation. $\newline$ $x = 7 \pm 4$
Two Possible Solutions: There are two possible solutions for $x$ . $x = 7 + 4$ or $x = 7 - 4$ $x = 11$ or $x = 3$

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