# 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, so the radius from the center of the circle to the point of tangency is perpendicular to the tangent line, as desired. $_\square$

## 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 $\Big(\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2}\Big)?$

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

There are two lines of slope $- \frac{2}{5}$ 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 $\frac{a \sqrt{b}}{c},$ 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 therefore

$PT^2=(r+AP)^2-r^2=AP \cdot (AP+2r),$

which is exactly what 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 the 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 to 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. $_\square$

## See Also

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

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