Explanation Given: End points of a diameter are ( 4 , 3 ) and ( 0 , 1 ) . Formula used: Distance formula: D(P ₁ , P ₂ ) = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Midpoint M = ( x , y ) = ( x 1 + x 2 2 , y 1 + y 2 2 ) Calculation: Let A ( x ₁ , y ₁ ) = ( 4 , 3 ) , B ( x ₂ , y ₂ ) = ( 0 , 1 ) Midpoint of a diameter of a circle is centre of the circle. Midpoint of AB = ( x , y ) = ( 4 + 0 2 , 3 + 1 2 ) = ( 4 2

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ISBN: 9780134119250

Author: Sullivan

Publisher: INTER PEAR

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Chapter 1.5, Problem 42SB

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To find: The standard form of equation of the circle with end points of a diameter at $(4, 3)$ and $(0, 1)$

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Chapter 1.5, Problem 41SB

Chapter 1.5, Problem 43SB

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The standard form of equation of the circle with end points of a diameter at ( 4 , 3 ) and ( 0 , 1 )

Solution Summary: The author explains the standard form of equation of the circle with end points of a diameter at ( 4, 3 ) and ( 0.

To find: The standard form of equation of the circle with end points of a diameter at $(4, 3)$ and $(0, 1)$

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Chapter 1 Solutions

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The standard form of equation of the circle with end points of a diameter at ( 4 , 3 ) and ( 0 , 1 )

Solution Summary: The author explains the standard form of equation of the circle with end points of a diameter at ( 4, 3 ) and ( 0.

To find: The standard form of equation of the circle with end points of a diameter at ( 4, 3 ) and ( 0, 1 )

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Chapter 1 Solutions

Additional Math Textbook Solutions

To find: The standard form of equation of the circle with end points of a diameter at $(4, 3)$ and $(0, 1)$