Two circles can generally be drawn. If the line joining the given points lie wholly without the given circle, one circle will

touch the given circle externally and one internally (as in the fig.); if the line joining the points cut the given circle, and both points lie on the same side of the circle, both circles will touch the given circle externally, and if the points lie on opposite sides of the circle both will touch it internally. If the line joining the given points touch the given circle one circle only can be drawn. Proof. The rectangle TM. TN rect. TB. TC (Euc. III. 36,

111.

Cor.), Euc. III. 36).

.. TD is a tangent to the circle going through B, D, C.

PROBLEM 28. (Fig. 27.) To describe a circle to touch a given circle (centre A, radius AR) and two given straight lines BC, DE.

There are several solutions depending on the relative positions of the lines and circle. If the lines are parallel the problem is impossible unless some part of the circle lies between the lines. In this case the line drawn midway between the lines parallel to either of them is evidently a locus of the required centre; a second locus will be the circle described with centre A and radius equal to the sum of AR and half the distance between the lines, and since these loci intersect in two points, either may be taken as

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CIRCLE TO TOUCH A CIRCLE AND TWO LINES.

the centre of the required circle. If the given lines are not parallel and the given circle cuts one of them, as in the fig., then by

drawing lines parallel to the given lines at a distance from them equal to the radius of the given circle, the problem may be reduced to describing a circle to touch these lines (in pairs) and to pass through the centre of the given circle, i.e. may be reduced to Problem 24. Let the given lines intersect in F and consider first the circles which can be drawn in the angle EFC. Draw GII, LK parallel to FE at a distance from it equal to AR and similarly HK and GL parallel to FC. FK will bisect the angle CFE and will be the locus of the required centres. Take any point 0 on it as centre and describe a circle to touch GH and GL cutting GA in M, M1. Then 40, drawn parallel to M,O to cut FK in 0, determines 0, a required centre and 40, parallel to MO determines 0, a 2 છુ ી second required centre. Similarly for the circles lying in the angle DFC. Any point 0, on FL being taken as centre and a circle described to touch GH and HK cutting HA in N and N1; 40, parallel to NO, determines 0,, the centre of a third circle fulfilling the required conditions and a line through A parallel to NO, would determine a fourth centre.

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It is of course accidental

If the given circle did not cut either of the given lines, it

would still be possible to draw four circles touching the lines and the circle, but two of them would have internal contact with the given circle, instead of all touching it externally as in the figure.

If the given circle cut both lines there would be six possible solutions.

PROBLEM 29. (Fig. 28.) To describe a circle to touch a given circle (centre A, radius AF) to touch a given line BC and to pass through a given point D.

If the given point be within the circle, the given line must not be wholly outside the circle.

From A draw AC perpendicular to the given line and meeting the circle in E and F. First join ED and on it determine a point

G such that the rect. ED. EG = rect. EC . EF, i. e. takę EG a fourth proportional to ED, EC, EF. [Making Ef (on ED) = EF, draw fg parallel to DC meeting EC in g and make EG-Eg.] Then a circle through Dand G and touching the given line will also touch the given circle and the problem is reduced to Problem 21. If ED, BC intersect in T, a mean proportional (TB) must be taken between TG and TD so determining the point of contact. TB may be set off along BC on either side of 7' and hence there are two solutions giving external contact. Second.-Join FD and on it determine a point G, such that rect. FD. FG1 = rect. on FC, FE. i.e. take a fourth proportional to FD, FC, FE. G, must be taken on the opposite side of F to D because C and E are on opposite sides of F. Then circles through D and G, touching the given line

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40 SEGMENT OF A CIRCLE TO CONTAIN A GIVEN ANGLE.

will also touch the given circle, and this case also reduces to Problem 21. There are again two possible circles because if DF and BC intersect in 7, the mean proportional (T,H) between TD and TG, may be set off on either side of T.

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Proof. Join B the point of contact of circle through D and G to E meeting the given circle in K and join FK. Then the triangles EKF and ECB are similar

or

but

.. EC: EB :: EK : EF,

rect. EC. EF-rect. EB. EK,

=rect. ED. EG (const.).

.. rect. ED. EG = rect. EB.EK,

.. K must be on circumference of circle through BDG. (Euc. III. 36 Cor.)

Join OK, then angle OBK = angle OKB (Euc. 1. 5),

and therefore OKA is a straight line, i.e. the two circles will touch at K.

PROBLEM 30. On a given straight line AB to describe a segment of a circle which shall contain a given angle (Fig. 29).

Bisect AB in C and through C draw CO perpendicular to AB. (CO is of course a locus of the centre.) Make the angle OCD

equal to the given angle (p. 4) and through 4 draw 40 parallel to CD meeting CO in 0. circle. (Euc. III. 20.)

O will be the centre of the required

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PROBLEM 31. (Fig. 30.) To draw a line touching two given circles, neither of which lies wholly inside the other.

A and AB are the centre and radius of the larger circle and C and CD those of the smaller circle.

Join AC, cutting larger circle in B.

From B on AC make BM BN = CD, and with A as centre describe circle MM,M,. From C draw tangents CM,, CM, to

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touch this circle (Prob. 20). Produce AM1, AM, to meet the circle in E and G and lines ED, GF through E and G parallel to CM, CM will be tangents to both circles. These tangents meet in (0,) a point lying on AC produced, and are the only pair that can be drawn if the given circles intersect. If the smaller circle lies wholly outside the larger, as in fig. 30, a second pair can be drawn by describing a circle through N with A as centre, drawing tangents CN1, CN to it, and drawing HJ, KL parallel to these lines re

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