# Find the locus of a point which moves so that the tangents from it to a circle are at right angles.

Find the locus of a point which moves so that the tangents from it to a circle are at right angles.

My Attempt : Let $P(x_1,y_1)$ be any moving point and $x^2 + y^2=a^2$ be the equation of the circle. Then, Centre of the circle is $(0,0)$ and its radius is $a$ Now, equation of tangent is $xx_1+yy_1=a^2$

How do I complete it?

You don't need analytic geometry to solve this problem: the following picture shows that the center of the circle, the two tangent points and point $P$ are the vertices of a square.

• Zar.Very nice ! – Peter Szilas May 21 '18 at 14:35

HINT

Consider one static position for that point $P$, then the distance from the center of the given circle is

• $d=\sqrt{a^2+a^2}=a\sqrt 2$

Now consider $P$ moving and note that distance $d$ can't varies in order to mantain the tangents orthogonal.

• what formula did you use for distance? Please elaborate – pi-π May 21 '18 at 14:15
• @blue_eyed_...consider a circle with radius "$a$" and two orthogonal tangents, then the intersction point P is the vertex of a square with side "$a$", then $OP=a\sqrt 2$ by Pythagoras. – gimusi May 21 '18 at 14:17
• could you please provide me a figure for this? – pi-π May 21 '18 at 14:19
• @blue_eyed_... refer to bing.com/images/… – gimusi May 21 '18 at 14:22

Let $Q,R$ be the points of tangency with the circle, $O$ the circle centre.

Note : $OQPR$ is a square, side length $= r$, $r$ is radius of the circle.

The locus of $P$ is a circle centred at $O$ with radius $R$:

$R =\sqrt{r^2+r^2} =√2r$\$