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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 $x y$ -plane. If $(x,-5)$ is a point on the circle, what is a possible value of $x$ ?

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Convert equations of conic sections from general to standard form

Full solution

Q.

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

\newline

The given equation represents a circle in the

x y

-plane. If

(x,-5)

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

x

?

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 $x$ . $(x-7)^2 + (-5+5)^2 = 16$
Simplify Equation: Simplify the equation by calculating $(-5+5)^2$ . $\newline$ $(x-7)^2 + (0)^2 = 16$ $\newline$ $(x-7)^2 + 0 = 16$ $\newline$ $(x-7)^2 = 16$
Eliminate Square: Take the square root on both sides of the equation to solve for $x$ . $\newline$ $\sqrt{(x-7)^2} = \pm\sqrt{16}$ $\newline$ $x - 7 = \pm4$
Solve for x: Solve for x by adding $7$ to both sides of the equation. $\newline$ $x = 7 \pm 4$
Calculate Possible Values: Calculate the two possible values of $x$ . $\newline$ $x = 7 + 4$ or $x = 7 - 4$ $\newline$ $x = 11$ or $x = 3$

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