Let's play pool

Pool

First the centre of the ball is marked as point B, the origin as O, and the centre of the cue ball at the time when it hits the ball be A.

Now if the cue ball is able to hit the ball into the hole then the neccasary and sufficient condition is $\angle ABO>{ 90 }^{ 0 }$

First in the triangle $ABO, AB=2r,OB=y$ and we assume $AO=x$

Applying cosine rule we have :

$cos(\angle BAO)=\frac { 4{ r }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 } }{ 4rx } <0$

$\Rightarrow { 4r }^{ 2 }+{ x }^{ 2 }<{ y }^{ 2 }\quad (i)$

Again applying cosine rule we have :

$cos(\angle OBA)=\frac { { 4r }^{ 2 }+{ y }^{ 2 }-{ x }^{ 2 } }{ 4ry }$

$\Rightarrow { x }^{ 2 }={ 4 }r^{ 2 }+{ y }^{ 2 }-4rycos\theta \quad (\theta =\angle OBA)\quad (ii)$

Using (ii) in (i) we have :

${ 4r }^{ 2 }+{ 4 }r^{ 2 }+{ y }^{ 2 }-4rycos\theta <{ y }^{ 2 }$

$\Rightarrow 2r<ycos\theta \quad (iii)$

(Refer to the diagram) In the right triangle BDE we have :

$cos(\angle DBE)=cos\theta =\frac { BD }{ BE }$

Put the values of BD and BE we get

$cos\theta =\frac { b-y }{ \sqrt { { a }^{ 2 }+{ (b-y) }^{ 2 } } }\quad(iv)$

Using (iv) in (iii) we get :

$2r\sqrt { { a }^{ 2 }+{ (b-y) }^{ 2 } } <y(b-y)$ (v)

$\Rightarrow ({ y }^{ 2 }-{ 4r }^{ 2 }){ (y-b) }^{ 2 }>{ 4r }^{ 2 }{ a }^{ 2 }$

Put the values to get :

$({ y }^{ 2 }-1){ (y-(\frac { 9+\sqrt { 5 } }{ 2 } )) }^{ 2 }>\frac { 15 }{ 2 } (3+\sqrt { 5 } )$

Solving for the inequality we have $y\epsilon (2,4)$ .

Hence $c=2,d=4$ hence $\boxed{{c}^{2}+{d}^{2}=20}$