What is the figure?

$(c+x), y, (c-x)$ are in GP $\Rightarrow y^2=c^2-x^2$ $\Rightarrow x^2+y^2=c^2$ Which is the equation of a circle having centre at origin & radius equal to c.

$(c+x),\quad y\quad and\quad (c-x)\quad are\quad in\quad G.P.\\ In\quad G.P,\\ \qquad { b }^{ 2 }=ac\\ Therefore,\\ \qquad { y }^{ 2 }=(c+x)(c-x)\\ \quad \quad \quad \quad \quad ={ c }^{ 2 }-{ x }^{ 2 }\\ So,\\ \qquad { x }^{ 2 }+{ y }^{ 2 }={ c }^{ 2 }\\ \\ Since\quad the\quad X\quad axis\quad and\quad Y\quad axis\quad are\quad perpendicular,\\ x\quad coordinate\quad and\quad y\quad coordinate\quad can\quad be\quad represented\\ as\quad base\quad and\quad perpendicular\quad of\quad a\quad right\quad angled\quad triangle.\\ \\ So\quad c\quad will\quad be\quad represented\quad as\quad the\quad distance\quad from\quad the\quad point\quad \\ to\quad the\quad origin.\\ As\quad c\quad is\quad constant,\quad the\quad graph\quad is\quad equedistant\quad from\quad the\quad origin.\\ \\ So\quad the\quad graph\quad is\quad a\quad \boxed { circle } \quad with\quad centre\quad (0,0)\quad and\quad radius\quad c.\\ \\ \qquad \qquad$

since C is constant, then the coordinates x and y having a common ration will make a fixed point and a point which moves an equal distance to the fixed point, thus forming a circle.