Find the equation of the circle whose center and radius are given. center ( 7, -3), radius = squareroot of 7

The answer is (x - 7)² + (y + 3)² = 7 or x² + y² - 14x + 6y + 51 = 0.
Solution:
We know that the standard form for the equation of a circle with the center located at coordinates (a, b) and radius r can be expressed as
     (x - a)² + (y - b)² = r²

Substituting the given information into the expression, we now have the equation of the circle whose center is at  (7, -3) and radius equal to square root of 7:
     (x - 7)² + (y + 3)² = (square root of 7)²
     (x - 7)² + (y + 3)² = 7

We can get the general form for the equation of the circle by expanding the equation of the standard form.
     (x - 7)² + (y + 3)² = 7
     (x - 7)(x - 7) + (y + 3)(y + 3) = 7
     x² -14x + 49 + y² +6y + 9 = 7
     x² + y² - 14x + 6y + 51 = 0

That is so true, i can’t believe how true this is