Griffin Spencer
01/27/2023 · Senior High School
17 In un sistema di assi cartesiani disegna la circonferenza di centro \( O(5 ; 6) \) e raggio uguale a 3 unità di misura. Fissa i punti \( A(2 ; 3) \), \( B(2 ; 6) \), \( C(4 ; 4), D(5 ; 9) \), \( E(9 ; 6) \). Specifica la posizio. ne di ciascuno di essi rispetto alla circonfe. renza.
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To solve this problem, we need to determine the position of each point relative to the circle with center \( O(5, 6) \) and radius 3 units. We will use the distance formula to calculate the distance between the center of the circle and each point, and then compare this distance to the radius of the circle.
The distance formula is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Where \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of two points.
Let's calculate the distance from the center of the circle to each point:
1. Point \( A(2, 3) \):
\[ d_A = \sqrt{(2 - 5)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.24 \]
2. Point \( B(2, 6) \):
\[ d_B = \sqrt{(2 - 5)^2 + (6 - 6)^2} = \sqrt{(-3)^2 + 0^2} = \sqrt{9} = 3 \]
3. Point \( C(4, 4) \):
\[ d_C = \sqrt{(4 - 5)^2 + (4 - 6)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.24 \]
4. Point \( D(5, 9) \):
\[ d_D = \sqrt{(5 - 5)^2 + (9 - 6)^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
5. Point \( E(9, 6) \):
\[ d_E = \sqrt{(9 - 5)^2 + (6 - 6)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4 \]
Now, let's determine the position of each point relative to the circle:
- Point \( A(2, 3) \): The distance \( d_A \approx 4.24 \) is greater than the radius of the circle (3 units), so point \( A \) is outside the circle.
- Point \( B(2, 6) \): The distance \( d_B = 3 \) is equal to the radius of the circle, so point \( B \) is on the circle.
- Point \( C(4, 4) \): The distance \( d_C \approx 2.24 \) is less than the radius of the circle, so point \( C \) is inside the circle.
- Point \( D(5, 9) \): The distance \( d_D = 3 \) is equal to the radius of the circle, so point \( D \) is on the circle.
- Point \( E(9, 6) \): The distance \( d_E = 4 \) is greater than the radius of the circle, so point \( E \) is outside the circle.
In summary:
- Point \( A \) is outside the circle.
- Point \( B \) is on the circle.
- Point \( C \) is inside the circle.
- Point \( D \) is on the circle.
- Point \( E \) is outside the circle.
Quick Answer
- Point \( A \) is outside the circle.
- Point \( B \) is on the circle.
- Point \( C \) is inside the circle.
- Point \( D \) is on the circle.
- Point \( E \) is outside the circle.
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