Circles and Construction Tools: Practice

Point $P$ is in the circle's plane, and its distance $5$ cm is positive and less than $8$ cm.

It is inside the disk, rather than on its boundary or at its center.

A diameter is twice the radius.

$18\div2=9$

The radius is $9$ cm.

Both $AO$ and $OB$ are radii, so each is $6$ cm. Since $O$ lies between the endpoints, segment addition gives

$AB=6+6=12.$

The diameter is $12$ cm long.

The endpoints lie on the circle, so $RS$ is a chord. It misses the center, so it fails the additional condition for a diameter.

The target point is on the circle centered at $P$ with radius $9$ cm.

Its distance from $P$ is therefore $9$ cm; changing the setting did not preserve the source length.

The copied chord uses the second setting, but $PE$ is a radius of the target circle centered at $P$.

That circle has radius $6$ cm, so $PE=6$ cm.

The two constructed sides are radii of the circles, so all three sides are $4$ cm long.

$4+4+4=12$

The perimeter is $12$ cm.

Halve the diameter to obtain the radius.

$24\div2=12$

The equation $x+4=12$ then gives $x=8$.

The point-on-circle condition gives

$2x+2=8.$

Subtract 2 and divide by 2 to obtain $x=3$.

Each copied segment has the original compass opening as its length. Segment addition gives

$9+9=18.$

Their combined length is $18$ cm.

Choose the intersection on the same specified side of the target baseline as the source opening. Here that is the upper intersection; the lower one places the opening on the other side.

The distance to one center equals that circle's radius, and the distance to the other equals the other radius. Different radii give different distances even when the point belongs to both circles.