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Pair of tangents drawn from a point $(x_1,y_1)$ to a circle $x^2+y^2=a^2$ is given by $$ SS_1=T^2\implies(x^2+y^2-a^2)(x_1^2+y_1^2-a^2)=(xx_1+yy_1-a^2)^2 $$

This is stated many where but most do not explain where does this come from. It is asked in a similar post The equation of a pair of tangents to a circle from a point, but is it possible to get an intuition into how the expression gives pair of tangents drawn from a point $(x_1,y_1)$ to a basic form of circle $x^2+y^2=a^2$ ?

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SOORAJ SOMAN
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1 Answers1

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

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

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

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

$$PQ\cdot SQ = \frac 12 (OP^2+OS^2-PS^2)-a^2=OP\cdot OS\cos\alpha-a^2$$

Employ the vector product to expression the above as,

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

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.

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