Circle Q is centered at the origin with radius r. Point P(x, y) lies on circle Q. Make a conjecture. How can you find an equation relating the radius to the coordinates of point P? Check all that apply. Notice that ΔPQS forms a right triangle.

Because ΔPQS is a right triangle,

apply the Pythagorean theorem.

x² + y² = r²

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JannetPalos JannetPalos · Answer 1 · 2018-12-19T13:20:14+01:00

Answer:

The relation [tex]x^{2} +y^{2} =r^{2}[/tex] is explained below.

Step-by-step explanation:

P(x, y) is a point on the circle. Q is the origin. Join PQ.

PQ is the radius.

Therefore, PQ = r

Draw PS perpendicular to the x-axis.

Now, ΔPQS is a right triangle.

By Pythagoros theorem,

[tex]PQ^{2}=PS^{2} +QS^{2}[/tex]

[tex]r^{2}=x^{2} +y^{2}[/tex]

[tex]x^{2} +y^{2} =r^{2}[/tex]

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abidemiokin abidemiokin · Answer 2 · 2022-01-10T14:32:50+01:00

An equation relating the radius to the coordinates of point P is [tex]r^2=x^2+y^2[/tex]

The square of the hypotenuse is equal to the sum of the square of the opposite and adjacent.

From the given diagram, we need to get the equation for the radius of the circle.

The radius of the circle is the line drawn from the center of the circle to its circumference. The radius is the hypotenuse side PQ

To get the radius of the circle, we will apply the Pythagoras theorem;

[tex]hyp^2=opp^2+adj^2\\r^2=x^2+y^2[/tex]

Hence an equation relating the radius to the coordinates of point P is [tex]r^2=x^2+y^2[/tex]

Learn more on Pythagoras theorem here: https://brainly.com/question/21970437