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CA; and draw CO at right angles to ED. The circle described with O as centre, and radius OA, will be the circle required.

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Since EA EC, the angle ECA= EAC, and ECO, EДO are equal, being right angles; ... OCA OAC and OA=

=

E

OC. The circle .. described from the centre O, and radius OA, will pass through the extremity of OC, and touch ED in C, because CO is at right angles to ED.

(37.) To describe a circle which shall pass through the extremities of a given line, so that if from any point in its circumference a line be drawn making a given angle with the given line; the rectangle contained by the segment it cuts off and the given line, may be equal to the square of the line drawn from the same point to the farther extremity of the given line.

Let AB be the given line. On it describe a segment of a circle containing an angle equal to the given angle. Complete the circle; it will be the one required.

BD

From any point C draw CD, making with AB the angle ADC equal to the given angle; join CA, CB. Since the angle CDA=ACB, and the angle at A is common, the triangles ACD, ABC are equiangular, and therefore

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whence the rectangle contained by AB, AD, is equal to the square of AC.

(38.) To determine a point in the perpendicular let fall from the vertical angle of a triangle on the base; about which as a centre a circle may be described touching the longer side, and passing through the opposite angular point.

Let ABC be a triangle, and from B the vertex let BD be drawn perpendicular to AC. In DB take any point E, and from it draw EF perpendicular to AB; and from E to BC, draw EG=EF;

B

F

E

H

D

from C draw CH parallel to GE, and from H draw HI perpendicular to AB; H is the point required.

Since EF is parallel to HI,

FE HI: BE : BH,

and since GE is parallel to HC,

GE

HC:: BE : BH,

.. FE HI :: GE HC;

but, by construction, FE=EG,.. HI=HC; and a circle described from the centre H at the distance HI, will pass through C, and touch AB in I, since the angle HIB is a right angle.

(39.) To describe a circle which shall have a given radius, and its centre in a given straight line, and shall also touch another given straight line inclined at a given angle to the former.

Let AB be the given line, in which the centre is to be; BC the line which the circle is to touch.

In BC take any point C, and draw CD at right angles to it; and make

с

D

B

CD equal to the given radius. Through D draw DO parallel to CB; O is the centre of the circle required.

Through O draw OE parallel to DC; .. CO is a parallelogram; whence OE is equal to DC, i. e. to the given radius. With the centre O, and radius OE, describe a circle; it will touch CB in E, because CO being a parallelogram, and ECD a right angle, CEO is also a right angle.

(40.) To describe a circle, which shall touch a straight line in a given point, and also touch a given circle. See (35)

Let AB be the given line, and C the given point in it, O the centre of the given circle. Draw CD perpendicular to AB, and OE parallel to CD. Join CE, meeting the circumference in F. Join OF, and produce it to meet CD in D. D is the centre of the circle required.

E

F

D

B

Since the triangles OEF, CFD are similar, and OE =OF, .. FD=DC; consequently a circle described with the centre D, and radius DF, will pass through C, and touch AB in C, because the angles at C are right angles; and it will touch the given circle in F, since the line joining the centres passes through F.

(41.) To describe two circles, each having a given radius, which shall touch each other, and the same given straight line on the same side of it.

Let AB be the given straight line. From any point A in it, draw AC at right angles to it, and make AC, AD, equal to the given radii. Produce CA to E, making AE=AD. Draw DO

parallel to AB; and with the centre C,

D

and radius CE, describe a circle cutting DO in O. C and O will be the centres of the circles required.

Join CO; and draw OB perpendicular to AB; then DAB being a right angle, as also ABO, .. AD is parallel to BO; and DO was drawn parallel to AB, :. AO is a parallelogram, and OB=AD. With the centres Cand O, and radii CA, OB describe circles, they will touch AB, since the angles at A and B are right angles; they will also touch each other, for CO is equal to CE, or to CA and AE, i. e. to CA and AD, or the sum of the radii.

(42.) To describe a circle passing through two given points, and touching a given circle.

Let A and B be the given points, and CDE the given circle..

circle through A and B, and

Describe a

cutting the

E

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given circle in

D and E. Join DE,

Then the angle EDA

EB, DA, AB.

=EBA; if .. DE and BA be pro

duced to meet in F, the triangle FDA will be similar to the triangle FBE;

and.. DF: FA :: BF : FE,

or the rectangle DF, FE is equal to the rectangle AF, FB. Draw FG a tangent to the given circle; then the

square of FG is equal to the rectangle EF, FD, and .. to the rectangle BF, FA; whence a circle described through the points A, G, B, will touch the given circle, since it touches FG.

(43.) To describe a circle, which shall pass through a given point, and touch a given circle and a given straight line.

Let ABC be the given circle, D the given point, and EF the given straight line. Through O draw AOE perpendicular to EF. Join AD; and divide it in G, so that the rectangle AG, AD, may be equal to the rectangle AC, AE. Through

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F

G and D describe a circle touching EF in F; this will also touch the circle ABC.

Draw the diameter FH; it is (Eucl. iii. 18.) parallel to AE. Join AF, meeting the circle in B. Join CB. The triangles ABC, AEF having the angle at A common, and the angles ABC, AEF right angles, are similar; whence

AC AB AF : AE,

.. the rectangle AB, AF is equal to the rectangle AC, AE, i. e. to the rectangle AG, AD; .. B is a point in the circle HDF. Take I the centre; join OB, BI. Since AC is parallel to FI, the angle OAB=BFI; but OAB=OBA, and IFB = IBF, .. OBA=IBF; and OBI is a straight line, which joins the centres of the two circles, which.. touch each other.

D D

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