Let the straight line $y = 2 x$ touch a circle with center $( 0 , \alpha ) , \alpha > 0$, and radius $r$ at a point $A _ { 1 }$. Let $B _ { 1 }$ be the point on the circle such that the line segment $A _ { 1 } B _ { 1 }$ is a diameter of the circle. Let $\alpha + r = 5 + \sqrt { 5 }$.

Match each entry in List-I to the correct entry in List-II.

\textbf{List-I}

(P) $\alpha$ equals

(Q) $r$ equals

(R) $A _ { 1 }$ equals

(S) $B _ { 1 }$ equals

\textbf{List-II}

(1) $( - 2,4 )$

(2) $\sqrt { 5 }$

(3) $( - 2,6 )$

(4) 5

(5) $( 2,4 )$

The correct option is

(A) $( \mathrm { P } ) \rightarrow ( 4 )$, $( \mathrm { Q } ) \rightarrow ( 2 )$, $( \mathrm { R } ) \rightarrow ( 1 )$, $( \mathrm { S } ) \rightarrow ( 3 )$

(B) $( \mathrm { P } ) \rightarrow ( 2 )$, $( \mathrm { Q } ) \rightarrow ( 4 )$, $( \mathrm { R } ) \rightarrow ( 1 )$, $( \mathrm { S } ) \rightarrow ( 3 )$

(C) $( \mathrm { P } ) \rightarrow ( 4 )$, $( \mathrm { Q } ) \rightarrow ( 2 )$, $( \mathrm { R } ) \rightarrow ( 5 )$, $( \mathrm { S } ) \rightarrow ( 3 )$

(D) $( \mathrm { P } ) \rightarrow ( 2 )$, $( \mathrm { Q } ) \rightarrow ( 4 )$, $( \mathrm { R } ) \rightarrow ( 3 )$, $( \mathrm { S } ) \rightarrow ( 5 )$