Write the standard equation of a circle that passes through (1, −6) with center (7, −2).

A: (x + 1)2 + (y – 6)2 = 52

B: (x + 7)2 + (y – 2)2 = 52

C: (x – 1)2 + (y + 6)2 = 52

D: (x – 7)2 + (y + 2)2 = 52

D: (x – 7)2 + (y + 2)2 = 52

Step-by-step explanation:

Circle: (x -h)² + (y - k)² = r²    center (h , k)    r: radius

(1 , -6) on the circle: radius   √(1 - 7)² + (-6 - (-2))² = √36 + 16 = √52

(x -7)² + (y +2)² = (√52)² = 52

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

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