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If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.?
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If two circles intersect at two points, prove that their centres lie o...
Statement: If two circles intersect at two points, their centers lie on the perpendicular bisector of the common chord.
Proof:
Step 1: Draw the diagram
First, draw two circles that intersect at two points. Label the centers of the circles as O1 and O2, and label the points of intersection as A and B.
Step 2: Connect the centers and draw the common chord
Next, connect the centers O1 and O2 with a straight line, and draw the common chord AB.
Step 3: Mark the midpoint of the common chord
Now, mark the midpoint of the common chord AB and label it as M.
Step 4: Prove that O1M is perpendicular to AB
To prove that O1M is perpendicular to AB, we need to show that the two lines are at right angles to each other.
Step 4.1: Prove that O1A is equal to O1B
Since O1 is the center of the first circle, the distances from O1 to A and O1 to B are equal. Thus, O1A = O1B.
Step 4.2: Prove that O1M is equal to O1M
Since M is the midpoint of AB, the distances from O1 to M and O2 to M are equal. Thus, O1M = O2M.
Step 4.3: Prove that triangle O1AM is congruent to triangle O1BM
From step 4.1 and 4.2, we know that O1A = O1B and O1M = O2M. Additionally, segment AM is equal to segment BM since M is the midpoint of AB. Therefore, triangle O1AM is congruent to triangle O1BM by the Side-Side-Side (SSS) congruence criterion.
Step 4.4: Prove that angle O1AM is equal to angle O1BM
Since triangle O1AM is congruent to triangle O1BM, the corresponding angles are equal. Therefore, angle O1AM is equal to angle O1BM.
Step 4.5: Conclude that O1M is perpendicular to AB
Since angle O1AM is equal to angle O1BM, and the sum of the angles in a straight line is 180 degrees, angle O1MA + angle O1MB = 180 degrees. Therefore, O1M is perpendicular to AB.
Step 5: Conclude that the centers lie on the perpendicular bisector
Since O1M is perpendicular to AB and M is the midpoint of AB, O1M is the perpendicular bisector of AB. Therefore, the centers O1 and O2 lie on the perpendicular bisector of the common chord AB.
Thus, it is proved that if two circles intersect at two points, their centers lie on the perpendicular bisector of the common chord.
Proof:
Step 1: Draw the diagram
First, draw two circles that intersect at two points. Label the centers of the circles as O1 and O2, and label the points of intersection as A and B.
Step 2: Connect the centers and draw the common chord
Next, connect the centers O1 and O2 with a straight line, and draw the common chord AB.
Step 3: Mark the midpoint of the common chord
Now, mark the midpoint of the common chord AB and label it as M.
Step 4: Prove that O1M is perpendicular to AB
To prove that O1M is perpendicular to AB, we need to show that the two lines are at right angles to each other.
Step 4.1: Prove that O1A is equal to O1B
Since O1 is the center of the first circle, the distances from O1 to A and O1 to B are equal. Thus, O1A = O1B.
Step 4.2: Prove that O1M is equal to O1M
Since M is the midpoint of AB, the distances from O1 to M and O2 to M are equal. Thus, O1M = O2M.
Step 4.3: Prove that triangle O1AM is congruent to triangle O1BM
From step 4.1 and 4.2, we know that O1A = O1B and O1M = O2M. Additionally, segment AM is equal to segment BM since M is the midpoint of AB. Therefore, triangle O1AM is congruent to triangle O1BM by the Side-Side-Side (SSS) congruence criterion.
Step 4.4: Prove that angle O1AM is equal to angle O1BM
Since triangle O1AM is congruent to triangle O1BM, the corresponding angles are equal. Therefore, angle O1AM is equal to angle O1BM.
Step 4.5: Conclude that O1M is perpendicular to AB
Since angle O1AM is equal to angle O1BM, and the sum of the angles in a straight line is 180 degrees, angle O1MA + angle O1MB = 180 degrees. Therefore, O1M is perpendicular to AB.
Step 5: Conclude that the centers lie on the perpendicular bisector
Since O1M is perpendicular to AB and M is the midpoint of AB, O1M is the perpendicular bisector of AB. Therefore, the centers O1 and O2 lie on the perpendicular bisector of the common chord AB.
Thus, it is proved that if two circles intersect at two points, their centers lie on the perpendicular bisector of the common chord.
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If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.?
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If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.?.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.?.
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