Director Circle

Locus of the point from which the two tangents to x²+y²=a² are perpendicular. A concentric circle with radius a√2. For circle (x−h)²+(y−k)²=r², the director circle is (x−h)²+(y−k)²=2r².

Let $P(h, k)$ be a point from which the two tangents to $x^2 + y^2 = a^2$ are perpendicular. Find the locus of $P$.

The combined equation of the pair of tangents from $(h, k)$ is $SS_1 = T^2$:

$$(x^2 + y^2 - a^2)(h^2 + k^2 - a^2) = (hx + ky - a^2)^2$$

For the two tangents to be perpendicular, the sum of the coefficients of $x^2$ and $y^2$ in the pair equation equals zero (standard condition for perpendicular lines from a pair).

Expanding and applying this condition leads to:

$$h^2 + k^2 = 2a^2$$

The locus is therefore:

$$x^2 + y^2 = 2a^2$$

Interpretation: The director circle is concentric with the original circle, with radius $a\sqrt{2}$ — a factor of $\sqrt{2}$ larger. Every point on the director circle sees the original circle at a right angle.

Alternative derivation: Let $OT_1$ and $OT_2$ be the two radii to the contact points. If the tangents are perpendicular, then $\angle T_1PT_2 = 90°$. The four points $O$, $T_1$, $P$, $T_2$ form a square (since $OT_1 \perp PT_1$ and $OT_2 \perp PT_2$), giving $OP = a\sqrt{2}$ directly.

For the circle $(x-h_0)^2 + (y-k_0)^2 = r^2$, the director circle is $(x-h_0)^2 + (y-k_0)^2 = 2r^2$.