We have to find the tangent to the circle x^2 + y^2 + 10x - 6y - 2 = 0 parallel to y = 2x without using calculus.

The given circle is

x^2+y^2+10x-6y-2=0

=> x^2+ 10x + 25 + y^2 - 6y + 9 = + 25 + 9 + 2

=> (x + 5) + (y - 3) = 6^2

The center of the circle is ( -5 , 3) and the radius is 6

Let the required line be y = 2x + c

The distance of (-5, 3) from 2x - y + c = 0 is 6

|2*(-5) - 3 + c| / sqrt ( 4 + 1) = 6

=> | -10 - 3 + c| = 6* sqrt 5

=> |-13 + c| = 6* sqrt 5

=> -13 + c = 6* sqrt 5

=> c = 6* sqrt 5 + 13

or 13 - c = 6* sqrt 5

=> c = 13 - 6* sqrt 5

**The equations of the tangents are 2x - y + 6*sqrt 5 + 13 = 0 and 2x - y + 13 - 6*sqrt 5 = 0**