Radical Triangles!

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There are three circles $C_1,C_2,C_3$ given by the following respective equations:-

$(x+4)^2+y^2=4$ $x^2+(y-8)^2=1$ $(x-1)^2+(y-1)^2=1$

Let $L_{ij}$ be defined as the locus of a point $P$ such that tangents from $P$ on $C_i$ and $C_j$ have equal lengths.

Find the area of the triangle formed by $L_{12}, L_{23}$ and $L_{13}$

The answer is 0.

1 solution

Meet Udeshi
Dec 20, 2013

It is obvious that $L_{ij}$ is the radical axis of the two circles.

Consider the intersection point of $L_{12}$ and $L_{23}$ . Let the length of tangent from this point on $C_2$ be equal to $a$ . Because this point lies on $L_{12}$ , we can say that length of tangent from this point to $C_1$ will also be $a$ . Similarly we can also say that length of tangent to $C_3$ will be $a$ , because it also lies on $L_{23}$ .

Thus we have shown that the length of tangents from that point on $C_1$ and $C_3$ will be the same, hence this point should also lie on $L_{13}$ . This proves that the three lines are concurrent, and so the area will be 0.

Radical Triangles!

The answer is 0.

1 solution

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