Conics!

We need to find the area of the quadrilateral $PQRS$ which is trapezium. $Let\quad the\quad equation\quad of\quad the\quad common\quad tangent\quad be\quad y=mx+c.\\ Let\quad the\quad equation\quad of\quad circle\quad and\quad parabola\quad be\quad { x }^{ 2 }+{ y }^{ 2 }=\frac { { a }^{ 2 } }{ 2 } \quad and\quad y=4ax,\quad where\quad a=2.\\ Applying\quad condition\quad for\quad tangency\quad we\quad get,\\ =>\quad c=\frac { a }{ m } =\pm \frac { a }{ \sqrt { 2 } } \sqrt { 1+{ m }^{ 2 } } \\ =>\quad \frac { { a }^{ 2 } }{ { m }^{ 2 } } =\frac { { a }^{ 2 } }{ 2 } (1+{ m }^{ 2 })\\ =>\quad { m }^{ 4 }+{ m }^{ 2 }-2=0\\ On\quad solving\quad we\quad get\quad m=\pm 1,\\ \therefore \quad equation\quad of\quad tangents\quad will\quad be,\\ =>\quad y=x+a\quad and\quad y=-x-a\\ Now\quad we\quad can\quad find\quad the\quad intersection\quad points\quad of\quad tangent\quad and\quad circle\quad and\quad tangent\quad and\quad parabola.\\ \therefore \quad P(\frac { -a }{ 2 } ,\frac { a }{ 2 } )\quad Q(\frac { -a }{ 2 } ,\frac { -a }{ 2 } )\quad R(a,2a)\quad S(a,-2a)\\ Area\quad of\quad quadrilateral\quad PQRS=\frac { 1 }{ 2 } (PT)(PQ+RS)\\ =>\quad area=\frac { 1 }{ 2 } (a+\frac { a }{ 2 } )(a+4a)\\ =>\quad area=\frac { 1 }{ 2 } \times \frac { 3a }{ 2 } \times 5a\\ =>\quad area=\frac { 15{ a }^{ 2 } }{ 4 } \\ Putting\quad a=2,\quad area=\frac { 15 }{ 4 } \times 4=\boxed { 15 }$