How to Construct Tangents from a Point to a Circle

Given circle \(A(B)\) and point \(C\) outside the circle, construct the tangents from \(C\) to \(A(B).\)

Construction of the tangents from a point to a circle

Solution

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Copyright © 1996-2018 Alexander Bogomolny

Given circle \(A(B)\) and point \(C\) outside the circle, construct the tangents from \(C\) to \(A(B).\)

Construction of the tangents from a point to a circle

The construction includes drawing an auxiliary circle on \(AC\) as a diameter. The applet below illustrates the process.

Let \(D\) be the midpoint of \(AC\), then the circle in question is centered at \(D\), passes through \(A\) and \(C\), and will be denoted \((D)\). Circle \((D)\) meets the given circle \(A(B)\) in points \(E\) and \(F\). The angles \(AEC\) and \(AFC\) are inscribed in \((D)\) and subtend diameter \(AC\); both are, therefore, right.

But \(AE\) and \(AF\) are radii of \(A(B)\), and the line perpendicular to a radius of a circle at its end point is tangent to the circle. It follows that \(CE\) and \(CF\) are the sought tangents.

As a bonus, this construction shows how to construct a circle centered at a given point and orthogonal to a given circle. Indeed, cicle \(C(E)\) fits the bill, because it passes through \(E\) and \(F\) and has \(AE\) and \(AF\) as tangents.

Note that when point \(C\) is located on the radical axis of two circles, then a circle with center at \(C\) orthogonal to one of the given circles is automatically orthogonal to the other.


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|Contact| |Front page| |Content| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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