Stuck on quant question from today's class (20 Oct.)

Hi. Can someone please help me with the following question discussed in today’s class?

Finding the radius using the following equation:
x(x-10)+y(y-6)=2

Greg solved the equation but he had added a 25 to the x side and a 9 to the y side. Anyone knows why these two specific numbers were added?

so the equation of a circle is

(x - h)² + (y - k)² = r²

where (h,k) is the centre of the circle and r is the radius.

so we have to convert the equation x(x-10)+y(y-6)=2 in the form of (x - h)² + (y - k)² = r²

therefore,
∴ x(x-10)+y(y-6) = 2
∴ x² - 10x + y² -6y = 2
∴ x² -10x + 25 + y² -6y + 9 = 2 + 25 + 9
( this step is done since (x + y)² = x² + 2xy + y² so x² -10x = (x - 5)² -25 )

∴ (x - 5)² + (y - 3)² = 36
now this is in the form of (x - h)² + (y - k)² = r²

hence the equation of the circle above has radius 6 and centre at (5,3)