# Tangent - Perpendicular to Radius

A **tangent** to a circle is a line intersecting the circle at exactly one point, the **point of tangency** or **tangency point**. An important result is that the radius from the center of the circle to the point of tangency is perpendicular to the tangent line.

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## Proof

Let \(T\) be the point of tangency, \(O\) be the center of the circle, and \(P\) be the foot of the altitude from \(O\) to the tangency line. Suppose that \(P\) and \(T\) are different points.

Since \(\angle OPT = 90^{\circ}\) and \(OT < OP\), \(\angle OTP > \angle OPT\), so \(\angle OTP > 90^{\circ}\). But then \(\triangle OPT\) has an angle sum greater than \(180^{\circ}\), which is a contradiction. Thus \(P\) and \(T\) must be the same point, and so the radius from the center of the circle to the point of tangency is perpendicular to the tangent line, as desired.

## Finding the tangent line at a point

Given a circle and a point on the circle, it is relatively easy to find the tangent line using coordinate geometry. For example,

A circle of radius \(3\) is centered at the origin. What is the equation of the tangent line of the circle that passes through the point \((\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2})\)?

The slope of the line from the center of the circle (the origin) and the given point \((\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2})\) is \(-1\). Therefore, the slope of the tangent line is 1, and since the tangent line passes through the point \((\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2})\), the equation of the tangent line is \(y=x-3\sqrt{2}\).

There are two lines of slope \(- \dfrac{2}{5} \) that are tangent to a circle of radius \(1\) centered at the origin \((0,0)\). Only one of these lines has a positive y-intercept. Let the point of tangency of this line to the circle be \((x,y)\). When \(x+y\) is written in the form \(\dfrac{a \sqrt{b}}{c} \) where \(gcd(a,c)=1\) and \(b\) is not divisible by the square of any prime, what is \(a+b+c\)?

## Power of a point

The perpendicularity fact can also prove a special (but important!) case of power of a point. Suppose \(P\) is a point in the plane of a circle \(O\), so that \(PT\) is tangent to the circle at \(T\) and \(PO\) hits the circle at \(A\) and \(B\) (with \(PA<PB\)). Then \(\triangle OTP\) is right, so

\[OP^2=PT^2+OT^2 \implies (r+AP)^2=PT^2+r^2\] and so \[PT^2=(r+AP)^2-r^2=AP \cdot (AP+2r)\] which is exactly as Power of a Point predicts, since \(PB=PA+2r\).

## Common external tangent

The perpendicularity condition is particularly useful when dealing with multiple circles, as their common tangent must be perpendicular to *both* radii to the tangent points. This also implies that those two radii are parallel, so the tangent line, two radii, and line between the two centers form a trapezoid.

## Two externally tangent circles \(O_1\) and \(O_2\) have radii of 3 and 5 respectively. What is the length of their common external tangent?

Suppose the common tangent is tangent to the circle with radius 3 at \(T_1\) and the circle with radius 5 at \(T_2\). Let \(X\) be the foot of the altitude from \(O_1\) to \(T_2O_2\). Then \(T_1T_2=O_1X\), and \(O_1X^2+XO_2^2=O_1O_2^2\). But \(XO_2^2=T_2O_2-T_1O_1=5-3=2\), and \(O_1O_2=5+3=8\), so \(T_1T_2^2=8^2-2^2=60\). Therefore, \(T_1T_2=2\sqrt{15}\), which is the desired length.

## See also

**Cite as:**Tangent - Perpendicular to Radius.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/tangent-perpendicular-to-radius/