131. PROP. XXXIV. To find the radius, and the centre, of a given circle.

Sometimes a circle is said to be given, when the radius is known, because the radius alone is sufficient to fix the magnitude of the circle. It is not, however, in such case fully given, unless the position of the centre be also known. Here the circle is supposed to be given by being simply presented before our eyes, without either centre or radius marked upon it.

(1) Two methods of finding the centre are given in (50), both sufficiently practical. The first method may be applied to any arc, or segment, of a circle, as well as to a whole circle.

(2) Of course, when the centre is found, any straight line drawn from it and terminated by the circumference is the radius.

(3) If the given circle be such, that no straight lines can be drawn within it, as a circular fish-pond or the mouth of a coal-pit, the radius may still be found by taking certain measurements, as will be shewn in Part III. on Mensuration.

132. PROP. XXXV. From a given arc of a circle to cut off a part equal to another given smaller arc of the same circle, or of a circle with the same radius.

E

Let AB be a given arc from which it is required to cut off a part equal to CD, another given arc with the same radius. Join the chord CD; then with

A

B

the centre A, and radius equal to the chord CD, describe a small arc cutting the arc AB in E; the arc AE is the arc required.

=

For, drawing the chord AB, the chord AE chord AB; and equal chords in the same circle, or in circles of equal radii, subtend equal arcs (58); .. arc AE= arc CD.

133. PROP. XXXVI. To bisect a given arc of a circle.

Let A and B be the extreme points of the arc which it is required to bisect. With centres A and B, and any convenient radius greater than half the chord AB, describe two pairs of intersecting arcs on opposite sides of

the chord AB, and let C, D be the points of intersection. Join CD, meeting the arc AB in E; the arc AB is bisected in E.

For, joining AB, AE, BE, let CD meet the chord AB in F; then, by (101), CD bisects AB in F, and is at right angles also to it; .. the triangles AFE, BFE, are equal in

all respects (24); and .. the side AE= the side BE. But equal chords in the same circle subtend equal arcs (58); . arc AE=arc BE.

If the point C fall on the same side of the arc AB as D, then CD being joined, DC must be produced to meet the arc in E.

Or, if for want of room, or other cause, the two pairs of intersecting arcs cannot be drawn on opposite sides of the chord AB, they may be drawn on the same side in the manner pointed out in (101).

134. PROP. XXXVII. To construct a circle whose circumference shall pass through, 1st, two given points, and 2nd, three given points.

B

(1) Let A and B be two given points. With centres A and B, and any convenient radius describe two pairs of intersecting arcs; join the points of intersection, D, E, and any point in DE, or DE produced, being taken for the centre, if a circle be described A to pass through A, it will also pass through B.

For DE bisects AB at right angles (101), and therefore passes through the centre of any circle of which AB is a chord (49).

(2) Let C be a third given point; proceed as before with the two points B, C, drawing FG to bisect the line BC at right angles. Produce FG and DE, if necessary, till they intersect in O. With centre O, and radius OA, describe a circle, and it shall pass through A, B, and C.

For the centre of every circle, which can be made to pass through A and B, is in DE, or DE produced; and

the centre of every circle, which can be made to pass through B and C, is in FG, or FG produced. And the only point which these lines have in common is their point of intersection O; that is, the only circle which can at the same time pass through A and B, as well as B and C, is that which has O for its centre.

N. B. The only case in which this construction will fail is, when the three given points are in one and the same straight line. In that case the lines DE, FG will be parallel, and .. will never meet in a point 0.

COR. 1. Any number of circles* can be drawn through one given point, or two; but no more than one distinct circle can be drawn through three given points. Hence three points, given in the circumference of a circle, are sufficient to fix both the magnitude and position of the circle.

COR. 2. Hence, also, no two distinct circles can cut one another in more than two points. For, if they could cut one another in three points, then assuming those points for the three given points of the Proposition, two distinct circles would be drawn through them, which, by Cor. 1, is impossible.

135. PROP. XXXVIII. To draw a straight line touching a given circle, 1st, from a given point in the circumference, and 2nd, from a given point without it.

(1) Let O be the centre of the given circle†, and A

the given point in the circumference. Join OA, and draw AB at right angles to OA; AB touches the circle at the point A.

The proof is given in (55). (2) When the given point is without the circumference, apply the method given in (56).

N. B. The application of this

B

A

Prop. is of constant occurrence in Mechanical Drawing; and requires to be strongly impressed upon the beginner, because it appears, at first sight, so easy a matter, without any theory, to draw a straight line touching a given circle; whereas, out of the numberless straight lines,

Here, and in numerous other places, we use the word circle, for shortness, when we really mean the circumference of the circle (19). + The centre is either at once given, or may be found by (50).

which may be drawn through the given point, only one in the 1st case, and two in the 2nd case, will really touch the circle; and, so far from being easy, it is barely possi ble, to guess at a true tangent of a circle by the eye.

136. PROF. XXXIX.

To draw a tangent to a given circle which shall also be parallel to a given straight line.

B

F

D

Let AB be the given straight line; and find O the centre of the given circle. Draw OC perpendicular to AB, meeting AB in C, and A the circumference of the circle in D; through D draw EDF parallel to AB, and EDF shall be the line required.

Or, produce DO to meet the circumference in G, and through G draw HGI parallel to AB, or at right angles to DG; then HI is the line required.

E

H

For, since ACO is a right angle, and ED is parallel to AC,.. EDO is a right angle (34 Cor. 4); and .. ED touches the circle at D (55). Similarly, HG touches the circle at G, and is parallel to AB.

137. PROP. XL. To draw a tangent to a given circle which shall also make with a given straight line an angle equal to a given angle.

A

D

C H

E

B

Let AB be the given straight line; take any point C in AB, and from C draw CD making with AB an angle equal to the given angle (105). Find Ŏ the centre of the given circle, and draw OE perpendicular to CD, meeting the circumference of the circle in F. Through F draw GFH parallel to CD, or at right angles to OF; then GFH is the tangent required.

G

L

For, it touches the circle at F, since < OFG is a right angle (55), and, since it is parallel to CD, ‹ AHĞ= < ACD (34) = the given angle.

138. PROP. XLI. To draw a straight line which shall touch each of two given circles.

B

There will be two cases of this Prop. 1st, when the touching line is wholly on one side of the line joining the centres of the circles; and 2nd, when it crosses that line. Let 0, 0, be the centres of the two given circles, of which the greater has O for its centre, and, for the 1st case, with centre 0, and radius equal to the difference of the two given radii, describe a circle, and draw from the point o a straight line oA touching this circle in the point

கு

A (56); through A draw OAB meeting the circumference of the larger circle in B; through B draw Bb, parallel to oA, meeting the smaller circle in b; then Bb is the straight line required.

For, since OAo is a right angle (56), and Bb is parallel to oA, . also ≤ OBb is a right angle; and Bb touches the larger circle at the point B (55).

Again, joining oB and ob, since Bb is parallel to oA, and AB-ob, the triangles ABo, oBb are equal in all respects, and.. 40bB = 40AB= a right angle, .. Bb touches the smaller circle at b (55).

In the same manner another straight line may be drawn on the opposite side of Oo touching the two given circles, as Cc.

2nd case, when the common tangent is required to cross the line joining the centres of the circles.

With centre 0, and radius equal to the sum of the