7.2 Tangents Tiếp tuyến

$\overline{EB}$ is tangent at $E,$ so $\overline{OE} \perp \overline{EB}$ by Theorem 7-1. With the right angles at $O$ and $B,$ quadrilateral $OEBA$ has three right angles, so its fourth angle is $360^\circ - 3 \cdot 90^\circ = 90^\circ$ and $OEBA$ is a rectangle.

Opposite sides of a rectangle are equal, so $OA = EB = 13$ and $AB = OE.$ $\overline{OK}$ and $\overline{OE}$ are radii $r$:

Answer: $AB = 5.$

$\overline{CD}$ is tangent to the smaller circle at $T,$ so $\overline{OT} \perp \overline{CD}$ by Theorem 7-1, and $OT = OA = 8$ (radii of the smaller circle).

$OC = OD$ (radii of the larger circle), so the right triangles $OTC$ and $OTD$ are congruent by HL and $TD = \dfrac{CD}{2} = 15.$

$\overline{OB}$ and $\overline{OD}$ are radii of the larger circle; in right $\triangle OTD$:

Answer: $OB = 17.$

Tangents to a circle from a point are equal (the corollary of Theorem 7-1):

From $AB = 8$ and $AC = 12$:

Since $AE = AF$:

Answer: $x = 5.$

Let the circle have center $I$ and radius $r,$ and let it touch $\overline{AC}$ at $E$ and $\overline{BC}$ at $F.$ Tangents from a point are equal: $AE = AT = 3$ and $BF = BT = 10.$

$\angle IEC,$ $\angle IFC$ (Theorem 7-1) and $\angle C$ are right angles, so $IECF$ is a rectangle with $IE = IF = r,$ and $CE = CF = r.$ So $AC = 3 + r,$ $BC = 10 + r$ and $AB = 3 + 10 = 13.$

By the Pythagorean Theorem:

A radius is positive, so $r = 2,$ $AC = 5$ and $BC = 12$:

Answer: the perimeter is $30.$

$\overline{OW}$ and $\overline{OP}$ are radii of $\odot O,$ so $OP = OW = 7.$ The point of tangency $W$ lies on the line of centers, so $W,$ $O,$ $P$ are on one line and $PW = OW + OP = 14.$

$\overline{PW}$ and $\overline{PZ}$ are radii of $\odot P$:

Answer: $OZ = 21.$

Let the radii be $a,$ $b$ and $c.$ The circles are externally tangent, so the distance between two centers is the sum of the radii:

Adding the three equations gives $2(a + b + c) = 34,$ so $a + b + c = 17.$ Then:

Answer: the radii of $\odot A,$ $\odot B,$ $\odot C$ are $6,$ $3$ and $8.$

Theorem: the tangents at the ends of a diameter are parallel.

Given: $\overline{AB}$ is a diameter of $\odot O$; line $l$ is tangent at $A$ and line $m$ is tangent at $B.$ Prove: $l \parallel m.$

1. $l \perp \overline{AB}$ and $m \perp \overline{AB},$ by Theorem 7-1 (a tangent is perpendicular to the radius at the point of tangency, and $\overline{OA},$ $\overline{OB}$ lie on $\overline{AB}$).

2. In a plane, two lines perpendicular to the same line are parallel, so $l \parallel m.$

Given: line $m$ lies in the plane of $\odot P$ and $m \perp \overline{PZ}$ at $Z,$ where $Z$ is on $\odot P.$ Prove: $m$ is tangent to $\odot P.$

1. Suppose $m$ is not tangent to $\odot P.$ Since $Z$ is on both, $m$ meets $\odot P$ at a second point $W.$

2. $PW = PZ,$ because they are radii of $\odot P.$

3. $\overline{PZ} \perp m,$ so $\overline{PZ}$ is the shortest segment from $P$ to $m,$ and $PZ < PW.$

4. Statements 2 and 3 contradict each other, so the supposition is false: $m$ is tangent to $\odot P.$