Explanation Given: Given conditions are, x 2 + y 2 + D x + E y + F = 0 Center of circle is ( 2 , 8 ) and radius of circle is 3 . Definition: Write the equation for the standard form of circle as follow, ( x − h ) 2 + ( y − k ) 2 = r 2 Here, ( h , k ) is the centre of circle and r is the radius. Calculation: Consider the standard equation of circle, ( x − h ) 2 + ( y − k ) 2 = r 2 Substitute 2 for h , 8 for k and 3 for r in the above equation to satisfies the given condition,

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ISBN: 9781305191495

Author: Jerome E. Kaufmann; Karen L. Schwitters

Publisher: Cengage Learning US

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Chapter 8.CT, Problem 6CT

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The equation of circle that satisfies the given conditions are center at (2,8) and radius of circle is 3.

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Chapter 8.CT, Problem 5CT

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Chapter 8 Solutions

Intermediate Algebra

Ch. 8.1 - For Problems 110, answer true or false. The graph...Ch. 8.1 - Prob. 2CQ Ch. 8.1 - Prob. 3CQ Ch. 8.1 - Prob. 4CQ Ch. 8.1 - Prob. 5CQ Ch. 8.1 - Prob. 6CQ Ch. 8.1 - Prob. 7CQ Ch. 8.1 - Prob. 8CQ Ch. 8.1 - Prob. 9CQ Ch. 8.1 - Prob. 10CQ

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The equation of circle that satisfies the given conditions are center at ( 2 , 8 ) and radius of circle is 3.

Solution Summary: The author explains that the equation of the circle that satisfies the given condition is x2+y4x-16y+59=0.

The equation of circle that satisfies the given conditions are center at (2,8) and radius of circle is 3.

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Chapter 8 Solutions