To show that \( OC \) is perpendicular to \( AB \), we can use the properties of circles and chords.
Given:
- Chord \( AB \) is such that \( AC = CB \) (this means \( AB \) is bisected at point \( C \)).
- The radius \( r \) of the circle is 5 units.
- The length of \( AC \) is 3 units, which implies \( CB \) is also 3 units (since \( AC = CB \)).
- The distance from the center \( O \) of the circle to point \( C \) is 4 units.
Steps to Show \( OC \) is Perpendicular to \( AB \)
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Identify the Center and Points:
- Let \( O \) be the center of the circle.
- The points \( A \) and \( B \) are endpoints of the chord \( AB \), and \( C \) is the midpoint of \( AB \).
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Calculate Length of \( AB \):
The length of chord \( AB \) can be calculated as: \[ AB = AC + CB = 3 + 3 = 6 \text{ units} \] -
Use the Relationship in a Circle:
The perpendicular from the center \( O \) of a circle to a chord bisects the chord. Since \( C \) is the midpoint of \( AB \), if we can show that \( OC \) is perpendicular to \( AB \), we satisfy this property of the circle. -
Use Pythagorean Theorem:
In triangle \( OAC \) (where \( O \) is the center of the circle, and \( A \) is one endpoint of the chord):- \( OA \) is the radius and is equal to 5 units.
- \( OC \) is the distance from the center to the midpoint, which is given as 4 units.
- \( AC \) is half the length of the chord, which is 3 units.
We can use the Pythagorean theorem: \[ OA^2 = OC^2 + AC^2 \] Plugging in the values: \[ 5^2 = 4^2 + 3^2 \] \[ 25 = 16 + 9 \] \[ 25 = 25 \] This holds true.
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Conclusion:
Since the Pythagorean theorem confirms that \( OA^2 = OC^2 + AC^2 \), this means that \( OC \) is indeed perpendicular to \( AB \). This is because in triangle \( OAC \), we have established that \( O \) lies directly above \( C \) such that the radius \( OA \) forms a right triangle with \( OC \) and \( AC\).
Therefore, \( OC \) is perpendicular to chord \( AB \).