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A circle in the
xy-plane has its center on the line
x=3. If the point
(4,5) lies on the circle and the radius is
sqrt2, which of the following could be the center of the circle?
Choose 1 answer:
(A)
(3,3)
(B)
(3,4)
(C)
(3,5)
(D)
(3,7)

A circle in the $\newline$ xy-plane has its center on the line $\newline$ $x=3$ . If the point $\newline$ $(4,5)$ lies on the circle and the radius is $\sqrt{2}$ , which of the following could be the center of the circle? $\newline$ Choose $1$ answer: $\newline$ (A) $(3,3)$ $\newline$ (B) $(3,4)$ $\newline$ (C) $(3,5)$ $\newline$ (D) $(3,7)$

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Find trigonometric ratios using the unit circle

Full solution

Q. A circle in the

\newline

xy-plane has its center on the line

\newline

x=3

. If the point

\newline

(4,5)

lies on the circle and the radius is

\sqrt{2}

, which of the following could be the center of the circle?

\newline

Choose

1

answer:

\newline

(A)

(3,3)

\newline

(B)

(3,4)

\newline

(C)

(3,5)

\newline

(D)

(3,7)

Understand the problem: Understand the problem. $\newline$ We are given a circle with a radius of $\sqrt{2}$ and a point on the circle $(4,5)$ . The center of the circle lies on the line $x=3$ . We need to find which of the given options could be the center of the circle.
Use distance formula: Use the distance formula to find the distance between the center $(3,y)$ and the point $(4,5)$ . The distance formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , where $(x_1, y_1)$ and $(x_2, y_2)$ are points on the plane. Here, $(x_1, y_1)$ is the center of the circle $(3, y)$ and $(x_2, y_2)$ is the point on the circle $(4,5)$ .
Plug in values: Plug in the values into the distance formula. $\newline$ We know the radius is $\sqrt{2}$ , so the distance $d$ is $\sqrt{2}$ . Plugging in the values, we get $\sqrt{2} = \sqrt{(4 - 3)^2 + (5 - y)^2}$ .
Simplify equation: Simplify the equation. $\newline$ Squaring both sides to eliminate the square root gives us $2 = (1)^2 + (5 - y)^2$ .
Further simplify: Further simplify the equation. $\newline$ $2 = 1 + (5 - y)^2$ leads to $1 = (5 - y)^2$ .
Solve for y: Solve for y. $\newline$ Taking the square root of both sides gives us $1 = 5 - y$ or $y = 5 - 1$ , which simplifies to $y = 4$ .
Check the answer: Check the answer. $\newline$ The center of the circle must be on the line $x=3$ , and we found $y=4$ . So the center of the circle is $(3,4)$ , which is option $(B)$ .

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