 
 <prompt> ```Write an article based on this "Gather your digital or analog drawing tools. Start with one large circle. Create a smaller circle inside the original, tangent to one side. Draw an identical circle "across from" the smaller inside circle. Apply Descartes's Theorem to find the size of your next circles. Create your next set of circles. Carry on in this fashion to continue adding circles. For a challenge, try making a non-symmetrical Apollonian Gasket by changing the size of your second circle."

Article:``` <response>: For the steps wey dey below, we go make our own simple Apollonian Gasket. E fit be say you wan draw Apollonian Gaskets by hand or for computer. For any case, you go need to fit draw perfect round circles. Dis one dey very important. Since every circle for Apollonian Gasket dey perfectly tangent to the circles wey dey next to am, circles wey dey even slightly misshapen fit "throw off" your final product. If you dey draw the Gasket for computer, you go need program wey go allow you to easily draw circles of fixed radius from central point. Gfig, wey be vector-drawing extension for the free image editing program GIMP, fit be used, as well as plenty other drawing programs (see the materials section for relevant links). You go probably need calculator application and either word processor document or physical notepad for taking notes on curvatures and radii. For drawing the Gasket by hand, you go need calculator (scientific or graphing suggested), pencil, compass, ruler (preferably scale with millimeter markings, graph paper, and notepad for note taking. Your first task na easy one - just draw one big, perfectly round circle. The bigger the circle, the more complex your Gasket fit be, so try make the circle as large as your paper fit allow or as large as you fit easily see for one window for your drawing program. Next, draw another circle inside the first one wey dey smaller than the original, but still fairly large. The exact size of the second circle na up to you - there no correct size. However, for our purposes, make we draw our second circle so that e reach exactly halfway across our large outer circle. In other words, make we draw our second circle so that the central point na the midpoint of the large circle's radius. Remember say for Apollonian Gaskets, all circles wey dey touch dey tangent to each other. If you dey use compass to draw your circles by hand, recreate this effect by putting the sharp point of the compass for the midpoint of the large outer circle's radius, adjust your pencil so that e just touch the edge of the large circle, then draw your smaller inner circle. Next, make we draw another circle across from our first one. This circle suppose dey tangent to both the large outer circle and the smaller inner circle, wey mean say your two inner circles go touch for the exact midpoint of the large outer circle. Make we stop drawing for a moment. Now we don draw three circles for our Gasket, we fit use Descartes's Theorem to find the radius of the next circle we go draw. Remember say Descartes's Theorem na d = a + b + c ± 2 (sqrt (a × b + b × c + c × a )), where a, b, and c na the curvatures of your three tangent circles and d na the curvature of the circle wey dey tangent to all three. So, to find the radius of our next circle, make we find the curvature of each of the circles we don draw so far so that we fit find the curvature of the next circle, then convert this to its radius. Make we define the radius of our outer circle as 1. Because the other circles dey inside this one, we dey deal with its interior curvature (rather than its exterior curvature), and, consequently, we know say its curvature na negative. - 1/r = -1/1 = -1. The big circle's curvature na -1. The smaller circles' radii na half as large as the big circle's, or, in other words, 1/2. Since these circles dey touch each other and the large circle with their outside edge, we dey deal with their exterior curvature, so their curvatures na positive. 1/(1/2) = 2. The smaller circles' curvatures na both 2. Now, we know say a = -1, b = 2, and c = 2 for our Descartes's Theorem equation. Make we solve for d: d = a + b + c ± 2 (sqrt (a × b + b × c + c × a )) d = -1 + 2 + 2 ± 2 (sqrt (-1 × 2 + 2 × 2 + 2 × -1 )) d = -1 + 2 + 2 ± 2 (sqrt (-2 + 4 + -2 )) d = -1 + 2 + 2 ± 0 d = -1 + 2 + 2 d = 3. The curvature of our next circle na 3. Since 3 = 1/r, the radius of our next circle na 1/3. Use the radius value wey you just find to draw your next two circles. Remember say these circles go dey tangent to the circles wey you use for a, b, and c for Descartes's Theorem. In other words, they go dey tangent to both the original and second circles. For these circles to dey tangent to all three circles, you go need to draw them for the open spaces for the top and bottom of the area inside your large original circle. Remember say these circles' radii go be equal to 1/3. Measure 1/3 back from the edge of the outer circle, then draw your new circle. E suppose dey tangent to all three of the surrounding circles. Because they be fractals, Apollonian Gaskets dey infinitely complex. This one mean say you fit add smaller and smaller circles to your heart's content. You dey limited only by the precision of your tools (or, if you dey use computer, the ability of your drawing program to "zoom in"). Each circle, no matter how small, suppose dey tangent to three other circles. To draw each subsequent circle for your Gasket, plug the curvatures of the three circles wey e go dey tangent to into Descartes's Theorem. Then, use your answer (wey go be the radius of your new circle) to draw your new circle accurately. Note say the Gasket wey we choose to draw na symmetrical, so the radius of one circle na the same as the corresponding circle "across from am". However, know say not every Apollonian Gasket na symmetrical. Make we tackle one more example. Make we say say, after drawing our last set of circles, we now wan draw the circles wey dey tangent to our third set, our second set, and our large outer circle. The curvatures of these circles na 3, 2, and -1, respectively. Make we plug these numbers into Descartes's Theorem, setting a = -1, b = 2, and c = 3: d = a + b + c ± 2 (sqrt (a × b + b × c + c × a )) d = -1 + 2 + 3 ± 2 (sqrt (-1 × 2 + 2 × 3 + 3 × -1 )) d = -1 + 2 + 3 ± 2 (sqrt (-2 + 6 + -3)) d = -1 + 2 + 3 ± 2 (sqrt (1)) d = -1 + 2 + 3 ± 2 d = 2, 6. We get two answers! However, because we know say our new circle go be smaller than any of the circles wey e dey tangent to, only a curvature of 6 (and therefore a radius of 1/6) make sense. Our other answer, 2, actually refer to the hypothetical circle on the other side of the tangent point of our second and third circles. This circle dey tangent to both of these circles and to the large outer circle, but e go intersect the circles we don draw already, so we fit disregard am. All Apollonian Gaskets start the same - with a large exterior circle wey dey act as the edge of the fractal. However, there no reason say your second circle necessarily suppose get 1/2 the radius of the first - we just choose to do this above because e simple and easy to understand. For fun, try start a new Gasket with a second circle of a different size - this one go lead to exciting new avenues of exploration. After drawing your second circle (regardless of its size), your next act suppose be to draw one or more circles wey dey tangent both to am and to the large outer circle - there no right way to do this, either. After this, you fit use Descartes's Theorem to determine the radii of any subsequent circles, as shown above. 