7.1 Basic Terms Các thuật ngữ cơ bản

A point is inside the circle when its distance from $O$ is less than the radius and outside when it is greater; a figure that goes from inside to outside crosses the circle once.

a. The segment starts at $A$ inside and ends at $B$ outside, so it crosses the circle once and stops at $B$: $1$ point.

b. The line passes through $A$ inside the circle, so it leaves the circle once on each side of $A$: $2$ points.

Answer: a. $1$; b. $2.$

a. The ray starts at $B$ outside and passes through $A$ inside, so it crosses the circle once between $B$ and $A.$ It goes on beyond $A$ and leaves the circle again: $2$ points.

b. The ray starts at $A$ inside and passes through the center $O$; it leaves the circle once and never comes back: $1$ point.

Answer: a. $2$; b. $1.$

a. Draw two perpendicular diameters $\overline{AC}$ and $\overline{BD}.$ The diagonals of $ABCD$ are equal, bisect each other and are perpendicular, so $ABCD$ is a square, and its four vertices lie on the circle: possible.

b. Choose three points $A,$ $B,$ $C$ on the circle with $B$ close to $A$ and to $C$ on a short arc. Then $\angle B$ is obtuse, and the triangle is inscribed because its three vertices lie on the circle: possible.

Answer: a. possible; b. possible (drawings on paper).

a. Suppose parallelogram $ABCD$ is inscribed in a circle with center $O,$ and let its diagonals meet at $M$; the diagonals of a parallelogram bisect each other. $OA = OC,$ so $O$ lies on the perpendicular bisector of $\overline{AC},$ a line through $M.$ In the same way $O$ lies on the perpendicular bisector of $\overline{BD},$ also through $M.$

If $O \ne M,$ both bisectors would be line $OM,$ so $\overline{AC}$ and $\overline{BD}$ would both be perpendicular to $OM$ and parallel, but they meet at $M.$ So $O = M,$ the four half diagonals are radii, and $AC = BD.$

A parallelogram with equal diagonals is a rectangle, which has right angles: impossible.

b. Draw a diameter $\overline{AC},$ then choose $B$ on one side of it and $D$ on the other side, both on the circle: possible.

Answer: a. impossible; b. possible.

$OX = OY = 10$ (radii), so $\triangle XOY$ is isosceles. Draw the altitude $\overline{OM}$ to $\overline{XY}.$ The right triangles $OMX$ and $OMY$ are congruent by HL (equal hypotenuses, common leg $\overline{OM}$), so $XM = MY$ and $m\angle XOM = \dfrac{120^\circ}{2} = 60^\circ.$

$\triangle OMX$ is a 30-60-90 triangle with hypotenuse $OX = 10$; by Theorem 6-7:

Answer: $XY = 10\sqrt{3}.$

$OX = OY$ (radii), so $\triangle XOY$ is isosceles. The altitude $\overline{OM}$ to $\overline{XY}$ makes congruent right triangles $OMX$ and $OMY$ (HL), so $XM = \dfrac{15}{2}$ and $m\angle XOM = 60^\circ.$

In the 30-60-90 triangle $OMX,$ $XM$ is the longer leg and the radius $r = OX$ is the hypotenuse; by Theorem 6-7:

Answer: the radius is $5\sqrt{3}.$

1. Draw $\overline{OZ} \perp l$ with $Z$ on $l$ (if $l$ passes through $O,$ then $Z = O$ and $OZ = 0$).

2. $OZ \le OX,$ because the perpendicular segment from a point to a line is the shortest segment from the point to the line; $OX < 13,$ because $X$ is inside the circle. So $OZ < 13$ and $169 - (OZ)^2 > 0.$

3. On $l$ take $A$ and $B$ on opposite sides of $Z$ with $ZA = ZB = \sqrt{169 - (OZ)^2}$; they are two different points.

4. $\triangle OZA$ has a right angle at $Z,$ so by the Pythagorean Theorem $OA^2 = OZ^2 + ZA^2 = OZ^2 + 169 - OZ^2 = 169,$ and $OA = 13.$ In the same way $OB = 13$ (if $Z = O,$ then $OA = ZA = 13$ directly).

5. A point whose distance from the center equals the radius lies on the circle, so $A$ and $B$ lie on $\odot O$: line $l$ cuts $\odot O$ in two points.

Let $l$ be a line and $\odot O$ a circle with radius $r.$ Suppose $l$ meets $\odot O$ in three different points $A,$ $B$ and $C.$

Draw $\overline{OZ} \perp l$ with $Z$ on $l$ (if $l$ passes through $O,$ then $Z = O$). $OA = OB = OC = r,$ because all radii of a circle are equal.

By the Pythagorean Theorem in the right triangles $OZA,$ $OZB,$ $OZC$: $ZA^2 = r^2 - OZ^2,$ $ZB^2 = r^2 - OZ^2,$ $ZC^2 = r^2 - OZ^2,$ so $ZA = ZB = ZC.$

But on a line at most two points are at a given distance from $Z,$ one on each side of $Z.$ So two of the points $A,$ $B,$ $C$ would be the same point, which contradicts that they are different.

What we supposed is false, so a line intersects a circle in at most two points.