Locus of the circumcentre of a variable triangle.

Question

I have a question related to circle that "Given the circle with equation $x^2+y^2-2x-2y+1=0$. Tangents $PA$ and $PB$ are drawn from a point $P$ to the given circle.It is given that $P$ lies on the line $lx+my+n=0$, where $l$,$m$,$n$ are constants. Find the locus of the circumcentre of the triangle $PAB$."

My approach: We can take any point on the given line and easily find out the equation of chord of contact i.e $AB$ of tangents by the formula $T =0$. Now my question is how to proceed for the circumcentre. I know that circumcentre of a triangle is the point of intersection of perpendicular bisector of the sides of the triangle. Please provide any hint.

Answer 1

The locus is a straight line (dotted line in the figure below).

Let $C(1,1)$ be the center of the given circle. Let $E$, $F$ be the contact points of the tangents issued from $P$. Due to the symmetry with respect to line $PC$ of the two tangents, $PEF$ is an isosceles triangle; therefore, the center $G$ of the circumscribed circle, which is situated on the altitude issued from $P$ in triangle $PEF$ is aligned with $P$ and $C$.

Moreover, $G$ is the midpoint of segment PC for the following reason: radius $EC$ being orthogonal to tangent $PE$ in contact point $P$, $\widehat{PEC}$ is a right angle, thus $PC$ is a diameter of the circumscribed circle; therefore $G$ is the midpoint of $PC$.

The locus of such a midpoint is the homothetic line from the original line in the homothety $\frak{H}$$(C,1/2)$, thus a parallel line to the original line.

Remark: The equation of this line is necessarily of the form $\ell x + my +n_1=0$ because they share the same normal vector $(\ell,m)$ with the original line. It remains to find $n_1$. This can be done by using the distance formula from point $C$ to the two straight lines (https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line).

Answer 2

Hint. Let $C$ be the center of the given circle. Then the circumcircle of $PAB$ goes through $C$ and the triangles $\triangle PAC$ and $\triangle PBC$ are right triangles. This implies that the centre of the circumcircle is the midpoint of the segment $PC$. If follows that the locus of these circumcircles is a straight line which is parallel to the given one.

Answer 3

HINT

The given circle $C_1$ has unit radius and touches x-, y- coordinate axes at $(1,0),(0,1).$ Required variable circle center has diameter $P C_1$ with tangent lengths $ PA= PB= P C_1$ and its cicumcenter center is on the bisector of tangents $ PA,PB $ as center point of $PC_1$. (Corrected)