 Descartes said you could solve any geometric construction problem by translating a geometric construction problem into an algebra problem, solving the algebra problem, and then turning the algebraic solution into a geometric construction procedure. Now, Descartes had identified geometric construction procedures with arithmetic operations, so the third step is routine. The second step only involves the knowledge of algebra, so while the problem may be difficult or even impossible to solve, there are, in principle, no difficulties. And as always, the first step is the challenging one. How do you translate a geometric construction problem into an algebra problem? So one of the important construction problems is, given a curve, construct a tangent to the curve. So what's a tangent? So the one tangent that was defined in Euclidean geometry is the tangent to a circle. And this comes from the elements, Book 3, Proposition 16, the straight line drawn at right angles to the diameter of a circle from its end will fall outside the circle. And as a corollary to this, the tangent to a circle is perpendicular to the diameter at the point of tangency. And so Descartes had the following insight. In the appropriate circle, then use the tangent to the circle as the tangent to the curve. And that's a brilliant mathematical solution, because it turns one problem we can't solve into a different problem, which we still can't solve. So Descartes considers this problem. Suppose the circle meets the curve at a point. Now if the circle actually crosses the curve, the system of equations that correspond to the circle and the curve will actually have two solutions, corresponding to each of the intersection points. But what's called an osculating circle will have a double root at the point of contact. Descartes described this as the two solutions becoming one. And we might go so far as to call this Descartes' principle. The geometric property of tangency corresponds to the algebraic property of a repeated root. This allows Descartes to solve problems like find the osculating circle for a parabola. Descartes actually considered this problem for what we would call a horizontal parabola, but will illustrate Descartes' method using a more familiar parabola, y equals x squared. So our goal, let's try to find the osculating circle that touches a point p on the curve. So to do that, we'll draw a line py-ordinate y's and let the center of the circle be at k where ok is equal to s. Then let py equals x and oy equal y. That also means ky is s minus y. Now you'll notice that we have a right triangle here so the Pythagorean theorem applies. And so this length is s minus y. This length is x and this length, well, we'll call it r. And so we have s minus y squared plus x squared equals r squared. But since we're dealing with a circle, this value r has to be true for all points on the circle. And so all points on the circle must satisfy this equation. Now since py equals x squared, we have, and we can rearrange that a little. Now this is a quadratic equation and ordinarily it would have two solutions. But since we want this to have two equal solutions, we know that if a quadratic has two equal solutions y equals e, then it must have a factor of the form y minus e squared, which we can expand. And so that means the coefficient of y must be minus 2e. And so we know that p minus 2s is minus 2e and we can solve this for s. And so to have two equal solutions y equal to e, we require that s is equal to e plus p over 2. Now there's one last issue. As long as s is equal to e plus p over 2, then we'll have two solutions at y equal to e. What do we want y to be? Well since our goal is to find the oscillating circle at the point p, then our solution should be e equal to o y, which is what we're calling y. And so s is going to be y plus p over 2. And this amount here, s minus y, well that's p halves. And it's worth noting that p is from the equation of our parabola. It has nothing to do with either x or y. The center of the oscillating circle is p halves units above the line drawn ordinate y's regardless of where we are on the parabola. Or let's use Descartes method to find the tangent to the ellipse 9x squared plus 16y squared equals 25 at the point where x is equal to a. So if we graph the ellipse, we see that its major axis runs horizontally, so it might be easier to use a circle with center on the horizontal axis of the ellipse. As before, let p be the point on the ellipse and h the center of the oscillating circle. Let px be the line drawn ordinate y's and let o x equals x, px equals y, and o h equals s. So again there's a right triangle here so the Pythagorean theorem applies, and in fact because we have an oscillating circle, all points on the oscillating circle have to satisfy this equation. Now since every point p on the ellipse satisfies the equation of the ellipse, we can solve for either x squared or y squared, and we'll solve for y squared. Replacing this in our equation, doing a bunch of algebra, and again a quadratic will ordinarily have two solutions, but since we want this to have a factor of x minus e squared, then we need minus 2e, the coefficient of x, to be minus 32s over 7, the coefficient of x. And remember our goal is to find s, so we solve for s. So if s is 7 sixteenths e, then we have a repeated root at x equal to e. Since we want x equals a to be our solution, we let x equals a, and so s equals 7 sixteenths a. So in our ellipse, if o x equals a, and o h equals 7 sixteenths a, then h will be at the center of the oscillating circle.