(6.) If two circles cut each other, the straight line joining their two points of intersection, is bisected at right angles by the straight line joining their centres.

Join BD, DA, AC, CB.

Since AD=DB and DC is common to the triangles ADC, BDC and the base AC=CB, ... (Eucl. i. 8.) the angle ADE = BDE. Hence the two sides AD, DE are equal to the two BD, DE and the included angles are equal, .. (Eucl. i. 4.) AE = EB, and the angle DEA=DEB, and being adjacent, they are right angles, i. e. DC bisects AB at right angles.

(7.) To draw a straight line which shall touch two given circles.

1. If the circles be equal.

Let A and B be the centres, join AB; and from A

D

and B draw AC, BD at right angles to it; join CD. Then AC being parallel and equal to DB; CD is parallel to AB, :. CABD is a rectangular parallelogram; and the angles at C and D being right angles, CD is a tangent to both circles (Eucl. iii. 16. Cor.).

2. If the circles be unequal, and the line be required to touch them on the same side of the line joining the

centres.

Let A and B be the centres; join AB; and with the

centre B and distance equal to the difference of the given radii, describe a circle, and from A draw ДE touching it. Join BE and produce it to D, draw AC parallel to BD, and join CD.

Then AC being parallel and equal to DE, CD is equal and parallel to AE, .. ACDE is a parallelogram; and the angle AEB being a right angle, AED is also a right angle; hence the angles at C and D are right angles, and therefore CD touches both circles.

3. If the line be required to touch them on opposite sides of the line joining the centres.

With the centre B and radius equal to the sum of

E

B

the given radii describe a circle, to which from A draw a tangent AE. Join BE, and let it cut the given circle in D. Draw AC parallel to BE; join CD.

Then AC being equal and parallel to ED, ACDE is a parallelogram, and the angle AED being a right angle, the angles at C and D are right angles, and therefore CD touches both circles.

(8.) If a line touching two circles cut another line joining their centres, the segments of the latter will be to each other as the diameters of the circles.

Let the line AB touch the circles, whose centres are C and D, in A and B, and cut CD in the point E; CE will be to

ED in the ratio of the diameters of the circles.

E

B

Join CA, BD. Then the angles at A and B are right angles and the angles at E are vertically opposite, therefore the triangles AEC, BED are equiangular, and consequently

CE: ED :: CA: BD

:: 2 CA 2BD.

(9.) If a straight line touch the interior of two concentric circles, and be placed in the outer; it will be bisected at the point of contact.

Let AB touch the interior of two cir

cles, whose common centre is O, in the point C; AB is bisected in C.

Join OC; then (Eucl. iii. 18.) the

angles at Care right angles; and OC drawn

B

from the centre of the circle ADB at right angles to AB, bisects it (Eucl. iii. 3.).

(10.) If any number of equal straight lines be placed in a circle; to determine the locus of their points of bisection.

E

D

F

B

Let there be any number of lines AB, CD, placed in the circle whose centre is O, and let them be bisected in E, F; join OE, OF; then (Eucl. iii. 14.) these lines are equal, and therefore the locus will be a circle whose centre is O, and radius equal to the distance of the points of bisection from 0.

(11.) If from a point in the circumference of a circle, any number of chords be drawn; the locus of their points of bisection will be a circle.

From the given point A let any chord AB be drawn in the circle, whose centre is 0; bisect it in D. Join AO, BO and draw DE parallel to BO.

A

E

Then DE being parallel to BO, the triangles ADE, ABO are similar, and BO is equal AO, .. DE=EA; but AE AO: AD: AB (Eucl. vi. 2.), whence AE = 1⁄2 AO, .. ED = EA = { AO, and the locus will be a circle described on 40 as a diameter.

(12.) If on the radius of a given semicircle, another semicircle be described, and from the extremity of the diameters any lines be drawn cutting the circumferences, and produced so that the part produced may always have

a given ratio to the part intercepted between the two circumferences; to determine the locus of the extremities of these lines.

On AB the radius of the semicircle AEC let a semicircle ADB be described; and from A draw any line ADE, which produce till EF ED in the given ratio.

A

F

E

D

Produce AC to G, making CG: CB, in the given ratio, and join DB, EC, FG;

.. FE GC :: ED: CB :: DA : AB :: EA: CA, whence (Eucl. vi. 2.) FG is parallel to CE and DB, and the angle AFG is a right angle, and is in a semicircle whose diameter is AG; hence the locus required is a semicircle.

(13.) If from a given point without a given circle, straight lines be drawn, and terminated by the circumference; to determine the locus of the points which divide them in a given ratio.

Let A be the given point and BCD the given circle. Find O its centre and join AO, and divide it in E, so

B

C

E

H

that AO: AE in the given ratio, and find a point F, so that EF may be to OD in the given ratio, and with the centre E and radius EF describe a circle; it will be the locus required.

Draw any line AGC; join OC, EG. Since 40 : AE in a given ratio, as also OD: EF;

.. OC: EG :: AO : AE,