For suppose that the circle is described so as to touch the smaller given circle; let G be the point of contact; we have then to shew that the described circle will also touch the larger given circle. Join TG, and produce it to meet the larger given circle at H. Then the rectangle TG,TH is equal to the rectangle TC, TD (14); therefore the rectangle TG, TH is equal to the rectangle TE, TF and therefore the described circle passes through H.
Let O be the centre of this circle, so that OGA is a straight line; we have to shew that OHB is a straight line.
Let TG intersect the smaller circle again at K; then AK is parallel to BH(14); therefore the angle AKT is equal to the angle BHG; and the angle AKG is equal to the angle AGK, which is equal to the angle OGH, which is equal to the angle OHG. Therefore the angles BHG and OHG together are equal to AKT and AKG together; that is, to two right angles. Therefore OHB is a straight line.
Two solutions will be obtained, because there are two solutions of the problem in 10. Also, if each of the given circles is without the other, two other solutions can be obtained by taking for T the point between A and B where a straight line touching the two given circles meets B. The various solutions correspond to the circumstance that the contact of circles may be external or internal
16. To describe a circle which shall touch three given circles.
Let A be the centre of that circle which is not greater than either of the other circles; let B and C be the centres of the other circles. With centre B, and radius equal to the excess of the radius of the circle with centre B over the radius of the circle with centre A, describe a circle. Also with centre C, and radius equal to the excess of the radius of the circle with centre C over the radius of the circle with centre A, describe a circle. Describe a circle to touch externally these two described circles and to pass through A (15). Then a circle having the same centre as the last described circle, and having a radius equal to
