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(x-h)^2+(y-k)^2=r^2

Derive standard form of an equation of a circle with center C(5,2) and radius r=4, using the Pythagorean theorem.

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User Tesgoe
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Final answer:

The standard form of an equation of a circle with center C(5,2) and radius r=4 can be derived using the Pythagorean theorem. The equation is: x^2 + y^2 - 10x - 4y + 13 = 0.

Step-by-step explanation:

The standard form of an equation of a circle with center C(5,2) and radius r=4 can be derived using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. In this case, the equation is given as: (x-h)^2+(y-k)^2=r^2, where (h,k) represents the coordinates of the center of the circle and r represents the radius.

Substituting the given values of the center C(5,2) and the radius r=4 into the equation, we get: (x-5)^2 + (y-2)^2 = 16. Simplifying further, we have: x^2 - 10x + 25 + y^2 - 4y + 4 = 16. Combining like terms, the equation can be written in standard form as: x^2 + y^2 - 10x - 4y + 13 = 0.

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User Stato Machino
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