Exercise 1262Geometry › Plane geometryL1Added 4 Nov 2026

The radical axis bisects the common tangent

#power of a point#radical axis#tangent

Statement

Two circles intersect at two points $A$ and $B$ . A line touches the first circle at $T_1$ and the second circle at $T_2$ (a common tangent). The line $AB$ meets the line $T_1T_2$ at a point $M$ . Prove that $M$ is the midpoint of $T_1T_2$ .

Hint

Recall the power of a point: if a line through $M$ meets a circle at $X$ and $Y$ , then $MX\cdot MY$ is the same for all such lines, and equals $MT^2$ if $MT$ is tangent to the circle at $T$ .

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Solution

Idea. The point $M$ lies on the line $AB$ , which is a secant for both circles, and on the tangent line $T_1T_2$ .

Power of $M$ with respect to the first circle. The line $MT_1$ is tangent at $T_1$ , and $MAB$ is a secant through $A$ and $B$ . Hence
$MT_1^2=MA\cdot MB.$

Power of $M$ with respect to the second circle. In the same way, $MT_2^2=MA\cdot MB$ .

Conclusion. $MT_1^2=MT_2^2$ , so $MT_1=MT_2$ . As $M$ is on the line $T_1T_2$ , it is the midpoint of $T_1T_2$ . $\blacksquare$

Remark. The line $AB$ is the radical axis of the two circles: the set of points having equal power with respect to both. It is perpendicular to the line of centres.

Lesson. Whenever tangents and secants meet at a point, write the power of that point in two ways.

The radical axis bisects the common tangent

Statement

Solution

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