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

- Notice that [tex]$\triangle PQS$[/tex] forms a right triangle.
- Because [tex]$\triangle PQS$[/tex] is a right triangle, apply the Pythagorean theorem.
- [tex]$x^2 + y^2 = r^2$[/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's solve this step-by-step by considering the information given:

1. Understanding the Circle and Point:
- Circle [tex]\( Q \)[/tex] is centered at the origin [tex]\((0, 0)\)[/tex] with radius [tex]\( r \)[/tex].
- Point [tex]\( P \)[/tex] has coordinates [tex]\((x, y)\)[/tex] and lies on circle [tex]\( Q \)[/tex].

2. Form a Right Triangle:
- Imagine point [tex]\( P \)[/tex] at coordinates [tex]\((x, y)\)[/tex].
- From point [tex]\( P \)[/tex], draw a vertical line down to the x-axis, intersecting it at point [tex]\( S \)[/tex] with coordinates [tex]\((x, 0)\)[/tex].
- This creates right triangle [tex]\(\triangle PQS\)[/tex] where:
- [tex]\( PQ \)[/tex] is the hypotenuse (as it’s the vertical distance from [tex]\( P \)[/tex] to the origin),
- [tex]\( PS \)[/tex] is the vertical leg from [tex]\((x, y)\)[/tex] to [tex]\((x, 0)\)[/tex],
- [tex]\( QS \)[/tex] is the horizontal leg from [tex]\((x, 0)\)[/tex] to [tex]\((0, 0)\)[/tex].

3. Applying the Pythagorean Theorem:
- The Pythagorean Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Mathematically, this is written as:
[tex]\[ PQ^2 = PS^2 + QS^2 \][/tex]

- Here, hypotenuse [tex]\( PQ \)[/tex] is the radius [tex]\( r \)[/tex] of the circle.
- Vertical leg [tex]\( PS \)[/tex] is the vertical distance from [tex]\( P \)[/tex] to the x-axis, which is [tex]\( y \)[/tex].
- Horizontal leg [tex]\( QS \)[/tex] is the horizontal distance from [tex]\( Q \)[/tex] to [tex]\( S \)[/tex], which is [tex]\( x \)[/tex].

Substituting these into the Pythagorean Theorem gives:
[tex]\[ r^2 = y^2 + x^2 \][/tex]

4. Rewriting the Relationship:
- Rearranging this equation to make it clearer, we write:
[tex]\[ x^2 + y^2 = r^2 \][/tex]

This equation relates the radius of the circle [tex]\( r \)[/tex] to the coordinates of the point [tex]\( P(x, y) \)[/tex] on the circle. It confirms that [tex]\( x^2 + y^2 \)[/tex] equals [tex]\( r^2 \)[/tex], which is a fundamental property of any point on a circle centered at the origin.
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