Let LM be a given finite straight line, PAB a

straight line given in position, but indefinite in length, cutting the given circle ABC in A and B : It is required to find a point in PB, from which, if a tangent be drawn to the circle ABC, it shall be equal to L.

Produce (S. 73. 3.) AB to D so that AD X DB= LM'; and from the centre D, at a distance = LM, describe a circle cutting the circumference of the circle of ABC in C; draw DC .. DC=LM; .. but (constr.) AD × DB=LM'; .. AD × DB = DC'; ... (E 37. 3.) DC touches the circle ABC, in C; and (constr.) it is equal to LM, and is drawn from a point D in the given indefinite straight line PAB.

PROP. XCIV.

121. PROBLEM. To describe a circle that shall touch a given straight line, and that shall also touch two given circles.

Let AB and CD be the two given circles, and

PQ a given straight line; and first, let neither of the two given circles lie within the other: It is required to describe a circle which shall touch both the given circles AB and CD, and which shall also touch PQ.

Find (E. 1. 3.) the centres K and L of the circles AB and CD; and if the circles be unequal, let CD be the greater; from any semi-diameter, as LC, of the greater, cut off CF equal to a semi-diameter of the less circle; from the centre L, at the distance LF describe the circle FGE; from any point P, in PQ, draw (E. 11. 1.) PR ↑ to PQ, and make PR also equal to the semi-di

ameter of the less circle AB; through R draw (E. 31. 1,) RS parallel to PQ; describe (S. 92, 3.) a circle KHG, passing through the point K, touching RS, in H, and touching the circle EGF, in G; let I be the centre of the circle KHG; join K, I, and L, I, and I, H; .. (E. 18. 3.) the < IHR is a right ▲, and .. (constr. and E. 29. 1.) the exterior IMP is, also, a right ; and the figure PH is a ; .. (E. 34. 1.) MH PR; and (constr.) PR=KB or DG; .. MH = KB or DG; ... IB, IM and ID are all equal; and (E. 16. 3. cor. and S. 6. 3). a circle described from the centre I, at the distance IM, will touch PQ in M, the circle AB in B, and the circle CD in C.

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But if the two circles AB, CD, be equal to one another, find, as before, their centres K, and L, and draw RS at a perpendicular distance from PQ equal to the semi-diameter of AB or CD: Then, if (S. 88. 3.) a circle be described passing through K and L, and touching RS, it is evident, that its centre will be the centre of the circle which is to be described, and its semi-diameter will be found, as in the former case, by joining that centre and the centre of either of the two equal and given circles.

And, in a similar manner, the problem may be solved, when it admits of a solution, if the two given circles do not lie without one another.

122. PROBLEM.

PROP. XCV.

To describe a circle which shall touch a given circle, and pass through two given points, either both without the circle, or both within it.

Let A, B, be two given points, and CDE a given

circle: It is required to describe a circle which shall pass through A and B, and which shall also touch the circle CDE.

First, let the two given points, A and B, be without the circle CDE: And if A and B be equally distant from the centre of CDE, it is manifest (S. 6. 3.) that a circle described (S. 5. 1. cor.) so as to pass through the two given points, and

through the extremity of a diameter of the given circle drawn perpendicular to the straight line joining those points, will touch the given circle.,

But if the points, A and B, be not equally distant from the centre of the circle CDE, take any point F, without the circumference of CDE, and through A, B and F, describe (S. 5. 2. cor.) the circle AFB; draw (S. 79. 3.) HX, so that the straight lines which are drawn from any point of it, touching the two circles CDE, AGB, shall be equal to one another, and let BA, produced, meet HX in H; from H draw (E. 17. 3.) HC and HD, touching the circle CDE in C and D; lastly, describe (S. 5. 1. cor.) two circles, the one passing through B, A, and C, and the other through B, A, D; the circles so described shall touch the given circle CDE, in the points C and D, respectively.

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For, from H draw (E. 17. 3.) HG, touching the circle AGB in G: Then (E. 36. 3.) BH × HA= HG; but (constr.) HG=HC; .. BH × HA= HC2; ... (E. 37. 3.) HC touches the circle described through B, A and C; and (constr.) it also touches the circle CDE; .. (E. 3. def. 3.) the circle BAC which passes through A and B, touches the circle CDE in C.

In the same manner it may be shewn, that the circle described so as to pass through A, B and D, touches the circle CDE in D: And by a like construction may the problem be solved, when