Pair of tangent lines from a point $(x_1,y_1)$ to a circle $x^2+y^2=a^2$

Here is an intuitive derivation that is based on a geometric construction:

From the diagram, we have $PS^2=(PQ-SQ)^2$, or,

$$PQcdot SQ = frac 12 (PQ^2+SQ^2-PS^2)$$

With the right triangle relationships $PQ^2=OP^2-a^2$ and $SQ^2=OS^2-a^2$, we have,

$$PQcdot SQ = frac 12 (OP^2+OS^2-PS^2)-a^2=OPcdot OScosalpha-a^2$$

Employ the vector product to expression the above as,

$$PQcdot SQ=vec{OP} cdot vec{OS}-a^2tag{1}$$

Next, use the coordinates for $S(x,y)$ and $P(x_1,y_1)$ to represent the line segments in (1),

$$PQ^2 = x_1^2+y_1^2-a^2,>>>>>SQ^2 = x^2+y^2-a^2$$ $$vec{OP} cdot vec{OS} = xx_1+yy_1$$

Plug above coordinate expressions into (1) to get the tangent equations

$$(x^2+y^2-a^2)(x_1^2+y_1^2-a^2)=(xx_1+yy_1-a^2)^2 $$ where we have squared the equation to include both tangents.