Answer :
Sure! Let's solve the problem step by step.
We have a circle, Circle Q, centered at the origin (0, 0), and a point P(x, y) that lies on this circle. We want to find an equation relating the radius of the circle to the coordinates of point P.
### Step 1: Understand the Circle's Equation
For any circle centered at the origin with radius [tex]\( r \)[/tex], the standard equation is given by:
[tex]\[ x^2 + y^2 = r^2 \][/tex]
This equation helps us express the relationship between the coordinates of any point on the circle and the circle's radius.
### Step 2: Consider Points on the Circle
Since point P(x, y) lies on the circle, its coordinates must satisfy the circle's equation. In other words, substituting the coordinates of point P into the equation should hold true:
[tex]\[ x^2 + y^2 = r^2 \][/tex]
### Step 3: Consider the Geometry of the Situation
The hint given suggests noticing that triangle APQS forms a right triangle. In a right triangle like this, if you imagine line segments from the origin at Q to P, and another horizontal line to the x-axis, it forms a right triangle with:
- One leg along the x-axis (length [tex]\( x \)[/tex])
- Another leg from the x-axis to P (length [tex]\( y \)[/tex])
- The hypotenuse being the radius of the circle (length [tex]\( r \)[/tex])
### Step 4: Apply the Pythagorean Theorem
Because we have a right triangle formed, we apply the Pythagorean theorem, which states:
[tex]\[ \text{(leg}_1)^2 + \text{(leg}_2)^2 = \text{(hypotenuse)}^2 \][/tex]
Plugging in the segments we have:
[tex]\[ x^2 + y^2 = r^2 \][/tex]
### Conclusion
So the equation [tex]\( x^2 + y^2 = r^2 \)[/tex] describes the relationship between the radius of the circle and the coordinates of any point (x, y) on the circle. This is consistent with what we expect for a circle centered at the origin.
Thus, any time you have a point on a circle centered at the origin, the sum of the squares of its coordinates will equal the square of the radius of the circle.
We have a circle, Circle Q, centered at the origin (0, 0), and a point P(x, y) that lies on this circle. We want to find an equation relating the radius of the circle to the coordinates of point P.
### Step 1: Understand the Circle's Equation
For any circle centered at the origin with radius [tex]\( r \)[/tex], the standard equation is given by:
[tex]\[ x^2 + y^2 = r^2 \][/tex]
This equation helps us express the relationship between the coordinates of any point on the circle and the circle's radius.
### Step 2: Consider Points on the Circle
Since point P(x, y) lies on the circle, its coordinates must satisfy the circle's equation. In other words, substituting the coordinates of point P into the equation should hold true:
[tex]\[ x^2 + y^2 = r^2 \][/tex]
### Step 3: Consider the Geometry of the Situation
The hint given suggests noticing that triangle APQS forms a right triangle. In a right triangle like this, if you imagine line segments from the origin at Q to P, and another horizontal line to the x-axis, it forms a right triangle with:
- One leg along the x-axis (length [tex]\( x \)[/tex])
- Another leg from the x-axis to P (length [tex]\( y \)[/tex])
- The hypotenuse being the radius of the circle (length [tex]\( r \)[/tex])
### Step 4: Apply the Pythagorean Theorem
Because we have a right triangle formed, we apply the Pythagorean theorem, which states:
[tex]\[ \text{(leg}_1)^2 + \text{(leg}_2)^2 = \text{(hypotenuse)}^2 \][/tex]
Plugging in the segments we have:
[tex]\[ x^2 + y^2 = r^2 \][/tex]
### Conclusion
So the equation [tex]\( x^2 + y^2 = r^2 \)[/tex] describes the relationship between the radius of the circle and the coordinates of any point (x, y) on the circle. This is consistent with what we expect for a circle centered at the origin.
Thus, any time you have a point on a circle centered at the origin, the sum of the squares of its coordinates will equal the square of the radius of the circle.
