Write the standard form of the equation of the circle with center (2,-4) and radius the positive root of the equation x^2-5x-6=0.

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The required circle has center at (2, -4) and the radius is the positive root of x^2 - 5x - 6 = 0.

x^2 - 5x - 6 = 0

=> x^2 - 6x + x - 6 = 0

=> x( x- 6) + 1(x- 6) = 0

=>...

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The required circle has center at (2, -4) and the radius is the positive root of x^2 - 5x - 6 = 0.

x^2 - 5x - 6 = 0

=> x^2 - 6x + x - 6 = 0

=> x( x- 6) + 1(x- 6) = 0

=> (x + 1)(x - 6) = 0

x = -1 and x = 6

The positive root is 6 which is the radius.

The standard form of the circle is (x - 2)^2 + (y + 4)^2 = 6^2

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