Tangents and special configurations

So far we have studied tangents from a single point to a single circle. The next step is two circles and combinations of tangents and chords. These setups appear less frequently than the two basic theorems but are favourites for higher-mark exam questions and for diagram-rich proof problems.

Common tangents to two circles

Given two circles in a plane, how many common tangents do they have? The count depends on how the circles are positioned relative to each other.

Circles entirely outside each other (no intersection, no touching): 4 common tangents , 2 external and 2 internal.
Circles touching externally: 3 common tangents , 2 external and 1 at the point of contact.
Circles intersecting at two points: 2 common tangents (both external).
Circles touching internally: 1 common tangent at the point of contact.
One circle inside the other (no touching): 0 common tangents.

For two circles of centres $O_1, O_2$ and radii $r_1, r_2$ :

External tangent length (the tangent that does not cross the line $O_1O_2$ ): $\sqrt{d^2 - (r_1 - r_2)^2}$ , where $d = O_1O_2$ .
Internal tangent length (crosses $O_1O_2$ ): $\sqrt{d^2 - (r_1 + r_2)^2}$ . (Real only when $d > r_1 + r_2$ .)

Tangent at the point of contact of two circles

When two circles touch (externally or internally), the line joining the centres passes through the point of contact, and the common tangent at that point is perpendicular to the line of centres at that point.

Proof (external tangency). Let the circles touch externally at $P$ . Then by the tangent-radius theorem, the tangent at $P$ to the first circle is perpendicular to $O_1P$ ; the tangent at $P$ to the second circle is perpendicular to $O_2P$ . But $O_1, P, O_2$ are collinear (a fact you should justify in a board proof: the distance between centres is $r_1 + r_2$ , which is precisely $O_1P + PO_2$ if $P$ lies on the segment $O_1O_2$ ). So both perpendiculars are perpendicular to the same line $O_1O_2$ at the same point $P$ , meaning they coincide. Thus the common tangent at $P$ is unique. $\blacksquare$

Alternate segment theorem (preview)

(Not central to Chapter 10, but appears in some textbooks.) The angle between a tangent and a chord at the point of tangency equals the inscribed angle subtended by the chord in the alternate segment of the circle. This becomes important in Class XI; for the board exam at this level, just be aware it exists.

Tangent length to a circle from an external point

A handful of classic problems revolve around computing tangent lengths and chord-of-contact distances. The chief identity:

$\text{tangent length}^2 = (\text{distance from point to centre})^2 - r^2.$

When the circle is given by $(x - a)^2 + (y - b)^2 = r^2$ in coordinate geometry, the tangent length from $(x_0, y_0)$ is

$\sqrt{(x_0 - a)^2 + (y_0 - b)^2 - r^2}.$

Some classic configurations

Inscribed circle of a triangle. A triangle's three sides are tangent to a unique inscribed circle (incircle). The inradius is $r =$ (Area)/ $s$ , where $s$ is the semi-perimeter.

Tangents at the ends of a chord meet on the perpendicular bisector of the chord. The two tangents are equal in length (from the same external point if the chord doesn't pass through the centre), and the external point lies on the perpendicular bisector of the chord.

Angle in a semicircle is $90^\circ$ . If $AB$ is a diameter, then any point $P$ on the circle satisfies $\angle APB = 90^\circ$ . (Used heavily when a tangent meets a diameter line.)

Worked examples

Example 1. Two circles of radii $4$ and $9$ touch externally. The distance between their centres is:

$d = r_1 + r_2 = 13$ .

Example 2. Two circles of radii $5$ and $3$ have centres $10$ cm apart. Find the length of the common external tangent.

$\ell = \sqrt{d^2 - (r_1 - r_2)^2} = \sqrt{100 - 4} = \sqrt{96} = 4\sqrt{6}$ cm.

Example 3. Two circles of radii $5$ and $3$ have centres $10$ cm apart. Find the length of the common internal tangent.

$\ell = \sqrt{d^2 - (r_1 + r_2)^2} = \sqrt{100 - 64} = \sqrt{36} = 6$ cm.

Example 4. A triangle has sides $13, 14, 15$ . Find the inradius.

$s = 21$ ; area (by Heron) $= \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84$ ; $r = 84/21 = 4$ .

Example 5. Two tangents from an external point to a circle of radius $r$ meet at $T$ with $\angle ATB = 90^\circ$ . The chord of contact $AB$ has length:

In this case the kite $OATB$ is a square (since $\angle OAT = \angle OBT = 90^\circ$ , $\angle ATB = 90^\circ$ , hence $\angle AOB = 90^\circ$ ). So $OA = OB = AT = BT = r$ and $AB = r\sqrt 2$ .

Try it yourself

Two circles of radii $7$ and $3$ touch externally. Distance between centres?
Two circles of radii $6$ and $4$ have centres $20$ cm apart. External tangent length?
Internal tangent length for circles of radii $5$ and $4$ , centres $13$ cm apart?
A triangle has sides $7, 24, 25$ . Find the inradius.
Two circles of radii $4$ and $4$ touch externally. Number of common tangents?
Show that the line joining the centres of two externally touching circles passes through the point of contact.
From a point on the major arc of a circle, the angle subtended by a chord is half the angle subtended at the centre. Why? (State the inscribed angle theorem.)
Two tangents from an external point are perpendicular. The radius is $r$ . Find the chord of contact's length.
A circle of radius $r$ is inscribed in a right triangle with hypotenuse $c$ and legs $a, b$ . Show $r = (a + b - c)/2$ .
Two circles of radii $r$ and $R$ intersect at two points. Show the common chord is perpendicular to the line of centres.
A circle is inscribed in an equilateral triangle of side $a$ . Find the inradius in terms of $a$ .
Two unequal circles touch externally. Show that a third circle through the point of contact, tangent to one of the original circles, is tangent to the other (a special case of inversion).

Pitfalls / Insight

(a) Make sure to memorise the external vs internal common tangent formulas. The internal one involves $r_1 + r_2$ (since the tangent crosses the line of centres), the external involves $r_1 - r_2$ .

(b) For internal tangents to exist, the circles must be entirely outside each other: $d > r_1 + r_2$ .

(d) The "two tangents perpendicular at external point" configuration creates a square , a fact that converts a hard-looking problem into instant solution.