Search results
Results From The WOW.Com Content Network
Draw a circle centered at M through the point A. This is the Carlyle circle for x 2 + x − 1 = 0. Mark its intersection with the horizontal line (inside the original circle) as the point W and its intersection outside the circle as the point V. These are the points p 1 and p 2 mentioned above. Draw a circle of radius OA and center W. It ...
Gauss's circle problem asks how many points there are inside this circle of the form (,) where and are both integers. Since the equation of this circle is given in Cartesian coordinates by x 2 + y 2 = r 2 {\displaystyle x^{2}+y^{2}=r^{2}} , the question is equivalently asking how many pairs of integers m and n there are such that
where A 1 and A 2 are the centers of the two circles and r 1 and r 2 are their radii. The power of a point arises in the special case that one of the radii is zero. If the two circles are orthogonal, the Darboux product vanishes. If the two circles intersect, then their Darboux product is
Kissing circles. Given three mutually tangent circles (black), there are, in general, two possible answers (red) as to what radius a fourth tangent circle can have.In geometry, Descartes' theorem states that for every four kissing, or mutually tangent circles, the radii of the circles satisfy a certain quadratic equation.
The recursion terminates when P is empty, and a solution can be found from the points in R: for 0 or 1 points the solution is trivial, for 2 points the minimal circle has its center at the midpoint between the two points, and for 3 points the circle is the circumcircle of the triangle described by the points.
Circle packing in a square is a packing problem in recreational mathematics, where the aim is to pack n unit circles into the smallest possible square. Equivalently, the problem is to arrange n points in a unit square aiming to get the greatest minimal separation, d n , between points. [ 1 ]
The intersection points of this circle with the two given lines (5) are T1 and T2. Two circles of the same radius, centered on T1 and T2, intersect at points P and Q. The line through P and Q (1) is an angle bisector. Rays have one angle bisector; lines have two, perpendicular to one another.
Figure 9: The two tangent lines of the two tangent points of a given circle intersect on the radical axis R (red line) of the two solution circles (pink). The three points of intersection on R are the poles of the lines connecting the blue tangent points in each given circle (black). Gergonne's approach is to consider the solution circles in ...