It's as hard as 1-2-3

Let $AO=OB=r$ . Also, let $\angle APO=\alpha$ .

By Law of Cosines on $\triangle APO$ , we have $r^2=1^2+3^2-2\cdot1\cdot3\cdot \cos\alpha$

Now note that $\angle BPO=270-\alpha$

By Law of Cosines on $\triangle BPO$ , we have $r^2=2^2+3^2-2\cdot2\cdot3\cdot\cos(270-\alpha)$ .

Observe that $\cos(270-\alpha)=-\sin\alpha$ (draw it on the unit circle if you are not convinced).

Therefore, $r^2=2^2+3^2+2\cdot2\cdot3\cdot\sin(\alpha)$

Simplify, and we have a system of equations: $\left\{\begin{array}{l}r^2=10-6\cos\alpha \\ r^2=13+12\sin\alpha\end{array}\right.$

Setting the two equations equal: $10-6\cos\alpha=13+12\sin\alpha$

We want to work with either only $\cos\alpha$ or only $\sin\alpha$ , which makes us think of $\sin^2\alpha+\cos^2\alpha=1$ . Thus, we aim to square both sides of the equation.

Isolating the $\sin\alpha$ : $12\sin\alpha=-3-6\cos\alpha$

Dividing by $3$ and squaring: $16\sin^2\alpha=1+4\cos\alpha+4\cos^2\alpha$

Using $\sin^2\alpha=1-\cos^2\alpha$ : $16-16\cos^2\alpha=1+4\cos\alpha+4\cos^2\alpha$

Rearranging: $20\cos^2\alpha+4\cos\alpha-15=0$

Aha, a quadratic! Let's let $\cos\alpha=x$ for now. $20x^2+4x-15=0$

Solving using the quadratic equation, we get that $x=\dfrac{-1\pm 2\sqrt{19}}{10}$

But which one is the correct value of $x$ ? We know that there can only be a unique $\alpha$ , but there are two possible values of $x$ .

Let's observe that type of angle $\alpha$ is. $\alpha$ seems to be an obtuse angle. Therefore, $\cos\alpha < 0$ , and the value we want is $x=\dfrac{-1- 2\sqrt{19}}{10}$ .

So now we know that $\cos\alpha=\dfrac{-1- 2\sqrt{19}}{10}$ . Looking back at what we want to find, we see that we want to find $\pi r^2$ . We already have an equation for $r^2$ though: $r^2=10-6\cos\alpha$

Plugging in $\cos\alpha=\dfrac{-1- 2\sqrt{19}}{10}$ gives $r^2=10-6\left(\dfrac{-1- 2\sqrt{19}}{10}\right)=\dfrac{53+6\sqrt{19}}{5}$ .

Thus, the area of the circle is $\left(\dfrac{53+6\sqrt{19}}{5}\right)\pi$ . Clearly, $a=53$ $b=6$ $c=19$ $d=5$

Therefore, our answer is finally $53+6+19+5=\boxed{83}$ and we are finally done! $\Box$

Problem maker's note: This particular problem took me roughly 1.5 hours , the longest I have ever taken, from idea to finished product, so if you like it, please show your support!

Let $M$ be the midpoint of the $AB$ , whose length is $\sqrt {1^2+2^2}=\sqrt {5}$ . Since $OA=OB$ , therefore $OM\perp AB$ . Let $N$ denote the intersection of $AP,OM$ .

Since $\triangle NAM\sim \triangle BAP$ , by their corresponding side ratio equality we have $AM*AB=AN*AP\Rightarrow \frac {\sqrt {5}}{2}\sqrt {5}=1*AN=\frac {5}{2}$ $\Rightarrow PN=\frac {5}{2}-1=\frac {3}{2}$ .

At this point, one may be motivated to use the laws of cosine on $\triangle OPN$ since $cos\angle ONP=-cos\angle ANM=-cos\angle ABP=-\frac {2}{\sqrt {5}}$ , that is what I decided to do to find the length of $ON$ , which is denoted by $x$ :

$OP^2=9=PN^2+x^2-2xPNcos\angle ONP=\frac {9}{4}+x^2+\frac {6}{\sqrt {5}}x$

Solving this quadratic gives $x=\frac {3\sqrt {19}-6}{2\sqrt {5}}$ (of course..after omitting the negative).

By the similarity again we have $NM=AM*2=\sqrt {5}$ , hence $OM=ON+NM=\frac {3\sqrt {19}+4}{2\sqrt {5}}$ and now we can use the Pythagorean theorem to find the radius squared, which is $\frac {53+6\sqrt {19}}{5}$ . Since the area of a circle is pi*radius squared, this is the expression we need, thus our answer is $53+6+19+5=\boxed {83}$

Well I found a solution using Pythagoras' Theorem. Here it goes

Construction

Now we use Pythagoras' Theorem on the three right-triangles,

$r^2=y^2+\left(x+2\right)^2\;---(i)$

$r^2=x^2+\left(y+1\right)^2\;---(ii)$

$x^2+y^2=3^2\;\;\;---(iii)$

From $(i)$ and $(ii)$ we get $3+4x=2y\rightarrow\,y=\frac{3}{2}+2x\;---(iv)$

Using this in $(iii)$ we get a quadratic in $x$ which is $20x^2+24x-27=0$ . Taking the positive root(since $x>0$ ) we have $x=\displaystyle\frac{-6+3\sqrt{19}}{10}$ .

Also using $(iii)$ in any of $(i)$ or $(ii)$ we get $r^2=13+4x--(v)$

So area of circle $=\pi\,r^2=\left(13+4x\right)\pi=\left(\displaystyle\frac{53+6\sqrt{19}}{5}\right)\pi$ .

So $a+b+c+d=53+6+19+5=\boxed{83}$ .

Lets make the solution very simple but not tricky ....

For a triangle ABC you know the cosine rule-

$cos(A) =\frac{{AB}^{2} + {AC}^{2} - {BC}^{2}}{ 2.AB.AC}$

Now we will apply the cosine rule in triangle POA and in triangle POB , for angel P in both the triangles. Also note that OA and OB is radius of circle, say OA = OB = r

Applying cosine rule we have-

$cos(\angle OPA) = \frac{10 - {r}^{2}}{6} \Rightarrow \angle OPA = \cos^{-1}{\frac{10 - {r}^{2}}{6}}$

$cos(\angle OPB) = \frac{13 - {r}^{2}}{12} \Rightarrow \angle OPB = \cos^{-1}{\frac{13 - {r}^{2}}{12}}$

And clearly $\angle OPA$ + $\angle OPB$ = $\frac{3\pi}{2}$

If you remember $\cos ^{ -1 }{ x } + \cos ^{ -1 }{ y } = \cos ^{ -1 }{ (xy - \sqrt { (1 - { x }^{ 2 })(1 - {y}^{2})}}$

Now, if $\cos ^{ -1 }{ x } + \cos ^{ -1 }{ y } = \frac{3\pi}{2}$ $\Rightarrow$ $xy - \sqrt { (1 - { x }^{ 2 })(1 - {y}^{2})}$ = 0 $\Rightarrow$ ${ x }^{ 2 } + { y }^{ 2 } = 1$

Here x = $\angle OPA$ and y = $\angle OPB$ , So-

${ \left( \frac { 10 - { r }^{ 2 } }{ 6 } \right) }^{ 2 } + { \left( \frac { 13 - { r }^{ 2 } }{ 12 } \right) }^{ 2 } = 1$
$\Rightarrow$ $5({ r }^{ 4 })-106({ r }^{ 2 })+425=0$

Using Sridharacharya Formula ,

${ r }^{ 2 }=\frac { 106\pm \sqrt { 2736 } }{ 10 } \Rightarrow { r }^{ 2 }=\frac { 106\pm 12\sqrt { 19 } }{ 10 } \Rightarrow { r }^{ 2 }=\frac { 53\pm 6\sqrt { 19 } }{ 5 }$

Now its simple to compare and get that a = 53, b = 6, c = 19, d = 5 [Ignore negative sign, as demand of question]
$\Rightarrow$ $a + b + c + d = 83$

Assume that the given figure is true. Draw lines parallel to PA and PB through O. Mark distance of P from these lines, $x$ and $h$ . Make three equations, $h^{2} + x^{2} = 9$ ; $(h+1)^{2}+ x^{2}=r^{2}$ ; $h^{2}+ (x+2)^{2}=r^{2}$ . Solve to get the value of $r^{2}$ as $\frac { 6\sqrt { 19 } +53 }{ 5 }$ . And hence get $\boxed{83}$ , answer

A solution using the power of a point theorem (just Euclidean Geometry, no coordinate geometry, no trigonometry).

Let $OM \bot PA$ , $ON \bot PB$ , $PM = x$ , $PN = y$ .

$M$ is the midpoint of $AA'$ , hence $MA' = x + 1$ .

Similarly, $N$ is the midpoint of $BB'$ , hence $NB' = y + 2$ .

By power of a point theorem: $PA \cdot PA' = PB \cdot PB' \Rightarrow 1 \cdot \left( {2x + 1} \right) = 2 \cdot \left( {2y + 2} \right) \Rightarrow x = 2y + \frac{3}{2} \ \ \ \ \ (1)$

By Pythagorean theorem on $\vartriangle ONP$ : $O{N^2} + N{P^2} = O{P^2} \Rightarrow {x^2} + {y^2} = 9 \ \ \ \ \ (2)$ .

Combining (1) and (2) we get the quadratic equation $5{y^2} + 6y - \frac{{27}}{4} = 0$ , which gives the positive solution $y = \frac{{ - 6 + 3\sqrt {19} }}{{10}}$ .

Now, the power of point $P$ is given by the formula $h = O{P^2} - {R^2}$ , (where $R$ is the radius of the circle) and is opposite to the product $PB \cdot PB'$ , hence,

$\begin{gathered} O{P^2} - {R^2} = - PB \cdot PB' \Rightarrow {R^2} = 9 + 2 \cdot \left( {2y + 2} \right) \\ \Rightarrow {R^2} = 4y + 13 = 4 \cdot \frac{{ - 6 + 3\sqrt {19} }}{{10}} + 13 = \frac{{53 + 6\sqrt {19} }}{5} \\ \end{gathered}$

Thus, the area of the circle is $A = {R^2} \cdot \pi = \frac{{53 + 6\sqrt {19} }}{5} \cdot \pi$ , so $a + b + c + d = 53 + 6 + 19 + 5 =\boxed{83}$ .

Draw line $OA$ and $OB$ . Since $OB=OA=x$ and angle $OPA = 270 -$ angle $OPB$ , the law of cosines can be used. Let angle $OPA = a$ .

$3^2 + 1^2 - 2(3)(1)*cos(a) = 3^2 + 2^2 - 2(3)(2)*cos(270-a)$ . This simplifies to $3^2 + 1^2 - 2(3)(1)*cos(a) = 3^2 + 2^2 + 2(3)(2)*sin(a)$ . Let $A=cos(a)$ , then $sin(a) = sqrt(1-A^2)$ . By substitution, $9-6A = 13-12*sqrt(1-A^2$ ). By solving the quadratic, we substitute the value of $A$ into $3^2 + 1^2 - 2(3)(1)*A=x^2$ to find the square of the radius. By multiplying this by $π$ , we obtain an expression with $a$ , $b$ , $c$ , and $d$ . Then, $a+b+c+d=83$ .

OAB is an isosceles triangle, with a OA=OB=R the radius.
Let M be the midpoint of base AB, and so OM is the altitude on the base.
Let P(0,0) in coordinate axis. So A(0,-1), B(2,0) and M(1,-1/2).
Slope of AB is 1/2, so slope of OM is - 2. M is on OM so, OM is Y = - 2X + 3/2. $\\Since~ O~ lies~ on~ it~ as~ well~ as~ on~ circle ~X^2 + Y^2 = 3^2,\\ \implies~ X^2 + (- 2X + 3/2)^2 = 9.\\ Solving~ the~ quadratic~ and~ getting ~Y ~from~Y = - 2X + 3/2, we~ get,\\ X= \dfrac{6\pm 3\sqrt{19}}{10}~~and~~Y = \dfrac{3\mp 6\sqrt{19}}{10}.~~O~is~(X,Y)\\O~ is~to~left~and~ above~ P(0,0),~ we~ take ~ +sign~ for ~~Y,~ - for~ X.\\\therefore~R^2=OA^2=(X-0)^2 + (Y- [-1])^2\\= \left(\dfrac{6- 3\sqrt{19}}{10}\right )^2 +\left(\dfrac{3+ 6\sqrt{19}}{10} + 1 \right)^2\\= \dfrac{53+ 6\sqrt{19}}{5}\\\therefore a+b+c+d=\boxed{\color{#D61F06}{83}}$

we know that $OB = OA$ so what we are going to do is finding what is equivalent to each of them and do the math :D

first of all let's assume that $OPA = x$ and $OB = r$ and $OA = r$

in Triangle $OBP$ and from the cosine law

$r^2 = 13 - 12 cos(270-x)$ which is equal to

$r^2 = 13 + 12 sin(x)$

in Triangle $OPA$ and from the cosine law

$r^2 = 10 - 6 cos(x)$

so we can say that

$10-6cos(x) - 13 + 12 sin(x) = 0$

$3+6cos(x) + 12 sin(x) = 0$

we can assume that $t=tan(\frac{x}{2})$

then $sin(x) = \frac{2t}{1+t^2}$ and $cos(x) = \frac{1-t^2}{1+t^2}$

so

$\frac{3+3t^2}{1+t^2} + \frac{6-6t^2}{1+t^2} + \frac{24t}{1+t^2} = 0$

$\frac{3t^2-24t-9}{-1-t^2} = 0$ ... we multiply both sides by $-1-t^2$

$3t^2-24t-9 = 0$ divide by $3$ to get

$t^2-8t-3=0$

we solve and we get that

$t = 4 + \sqrt(19)$ and $x = 166.3559152$ and we know that

$r^2 = 13 + 12 sin(x)$ and it will be equal to

$r^2 = \frac{53+6 \sqrt(19)}{5}$

so the solution will be $35+6+19+5 = \boxed{83}$

and that is the first solution

$t=4-\sqrt(19)$ is the second solution but then $x$ will be acute which is refused and I don't know why ! :D ... I think this problem should have two answers

We choose $P$ to be the Origin of the plane and $A$ , $B$ to be $(0,1)$ , $(2,0)$ respectively. The mid point $M$ of $AB$ must be $(1, \frac{1}{2})$

We have $O$ is the intersection of $(d)$ - the line which passes through $M$ and perpendicular to $AB$ and $(P, 3)$ .

The equation of $(d)$ is $2x-y+\frac{3}{2}=0$ The equation of $(P,3)$ is $x^2+y^2=9$

To get the result, we calculate the squared of the radius of $(O)$ that is

$r^2=OB^2=(x_O-2)^2+y_O^2=x_O^2+y_O^2-4x_O+4=13-4x_O$

due to $O$ is the intersection

We can calculate $x_O$ by substitute $y_O=2x_O+\frac{3}{2})$ in $x_O^2+y_O^2=9$ , solve it and choose $x_O<0$ (look at the figure)

We have

$x_O=\frac{6-3\sqrt{19}}{10}$

and thus

$r^2=13-4x_O=\frac{53+6\sqrt{19}}{5}$

From the figure, we note that triangle AOB is isosceles such that AO=BO=x, where x is the radius of the circle.

∴ ∠OAP+∠PAB=∠ABO. Call ∠OAP α; ∠PAB β; and ∠ABO γ.

Our working inverse equation here is:

cos^(-1)⁡α+cos^(-1)⁡β=cos^(-1)⁡γ

From this, we can obtain that cos⁡(α+β)=cos⁡γ Also, applying the cosine law and trigonometric identity leads us to the following relations: cos⁡α=(x^2-8)/2x,cos⁡β=1/√5 ; sin⁡α=(2√(-x^4+20x^2-64))/2x, ⁡sin⁡β=2/√5 ; cos⁡γ=5/(2√5 x)

But cos⁡(α+β)=cos⁡γ is equivalent to cos⁡α cos⁡β-sin⁡α sin⁡β=cos⁡γ Plugging in the values, (x^2-8)/2x∙1/√5- (√(-x^4+20x^2-64))/2x∙ 2/√5=5/(2√5 x) (x^2-8)/(2√5 x)- (2√(-x^4+20x^2-64))/(2√5 x)=5/(2√5 x) x^2-13=-2√(-x^4+20x^2-64)

Squaring both sides, we have x^4-26x+169=4(-x^4+20x^2-64) 5x^4-106x^2+425=0 Let m be equal to x^2 such that 5m^2-106m+425=0 Getting the roots of m by applying the quadratic formula, m=(106±√(106^2-4(5)(425)))/(2(5)) Simplifying, m=(53±6√19)/5. The area of the circle can be inferred to as mπ. Therefore, a=53,b=6,c=19,d=5, and a+b+c+d=53+6+19+5= 83

Let us solve this problem with Co-ordinate Geometry let point P(0,0) A(0,-1), B(2,0) Now join AB and get the midpoint C(1,-0.5) Equation of AB: x-2y=2 Let the perpendicular line to AB is MC Equation of MC: 4x+2y=3 Now consider a circle Z with centre at P, x^2+y^2=9 Now the top intersecting point of Circle Z and MC is ( (3 (2 - Sqrt(19))/10) , (3 (1 + 2 Sqrt(19) )/10) ) Hence the intersecting point is O as "the the perpendicular line to the midpoint of any arc of a circle always passes through the centre of the circle" now we know the point O and from the question we know A & B from here we get Radius of the circle so the term [{a+bsqrt(c)}/d] is simply the term R^2 when R^2 we get [{53+6sqrt(19)}/5] so a=53, b=6, c=19, d=5 and a+b+c+d= 83

I took P to be origin and solved the equations.it will require less effort

Denote circumscribed circle of triangle by CCT, circle with center 0, radius a by (O, a). CCT(PAB) x (O, 3) = {P, E} $AB=\sqrt{5}, PE=AB-2(AP*cos(PAB))=AB-2(1*\frac{1}{\sqrt{5}})=\sqrt{(5)-2(1*1/\sqrt{5})}=3/\sqrt{5}$ $OE=\sqrt{3^2-(PE/2)^{2}}=\sqrt{9-9/20}=3*\sqrt{19/20}$ $R=\sqrt{(OE+2/\sqrt{5})^{2}+(\sqrt{5}/2)^{2}}=\sqrt{(3*\sqrt{19/20}+2/\sqrt{5})^{2}+(\sqrt{5}/2)^{2}}$ $R^{2}=(3*\sqrt{19/20}+2/\sqrt{5})^{2}+(\sqrt{5}/2)^{2}$ $=(3*\sqrt{19}+4)^{2}/20+5/4=171+16+24\sqrt{19}+25/20=(53+6\sqrt{19})/5$ $a+b+c+d=\boxed{83}$

Synthetic one (Without Trigonometry)

Suppose that $P_1$ and $P_2$ be points on lines $AP$ and $BP$ respectively such that $\angle{OP_1P = 90^{\circ}}$ and $\angle{OP_2P} = 90^{\circ}$ . Denote the lenght of $OP_1$ and $OP_2$ by $x$ and $y$ respectively and let the radii be $r$ , then, by simple application of pytagorean theorem, we have $OP_1^2 + OP_2^2 = x^2 + y^2 = OP^2 = 9 \cdots (1)$ $OA^2 = r^2 = OP_1^2 + P_1A^2 = x^2 + (1+y)^2 = x^2 + y^2 + 2y + 1 = 10 + 2y \cdots (2)$ $OB^2 = r^2 = OP_2^2 + P_2B^2 = y^2 + (2+x)^2 = x^2 + y^2 + 4x + 4 = 13 + 4x \cdots(3)$ So, we get $x = \frac{r^2-13}{4}$ and $y = \frac{r^2-10}{2}$ . Now, substitute these into $(1)$ : $\frac{r^4-20r^2+100}{4} + \frac{r^4-26r^2+169}{16} = 9$

After some manipulation, the equation turns into $5r^4 - 106r^2 + 425 = 0 \longrightarrow r^2 = \frac{106 + 12\sqrt{19}}{10} = \frac{53 + 6\sqrt{19}}{5}$ So, $a + b + c + d = 53 + 6 + 19 + 5 = \boxed{83}$ .

Doing\quad it\quad with\quad co-ordinate\quad geometry\quad is\quad much\quad simpler.\ Take\quad O\quad as\quad origin\quad i.e.(0,0)\ P\quad as\quad (x,y)\ A\quad as\quad (x,y-1)\ B\quad as\quad (x+2,y)\ Given\quad PO\quad =\quad 3,\quad \ \dot { .\quad . } \quad { x }^{ 2 }+{ y }^{ 2 }=9\quad ...(1)\quad \quad \quad \quad \quad \quad \quad (Distance\quad formula)\ OA=OB\quad (Radius\quad of\quad the\quad circle)\ \dot { .\quad . } \sqrt { { x }^{ 2 }+{ (y-1) }^{ 2 } } =\sqrt { (x+2)^{ 2 }+{ y }^{ 2 } } \quad (Distance\quad formula)\ Squaring\quad both\quad sides\quad and\quad solving\quad further\quad we\quad get\ 4x+2y=-3\quad ...(2)\ Substituting\quad y=\frac { -3-4x }{ 2 } \quad in\quad (1)\quad we\quad get\quad a\quad quadratic\quad eq\quad in\quad terms\quad of\quad x\ i.e.\quad 20{ x }^{ 2 }+24x-27=0\ whose\quad roots\quad are\quad \frac { -6+3\sqrt { 19 } }{ 10 } \quad and\quad \frac { -6-3\sqrt { 19 } }{ 10 } .\ Discarding\quad the\quad second\quad root\quad since\quad x\quad needs\quad to\quad be\quad positive\quad as\quad it\quad lies\quad to\quad right\quad side\quad of\quad the\quad origin\quad O.\ Now\quad we\quad know\quad Radius\quad r\quad =\quad OB.\ { r }^{ 2 }\quad =\quad (x+2)^{ 2 }+{ y }^{ 2 }\ { r }^{ 2 }\quad =\quad { x }^{ 2 }+4x+4+{ y }^{ 2 }\ { r }^{ 2 }\quad =\quad 13+4x\quad \quad \quad \quad (Since\quad { x }^{ 2 }+{ y }^{ 2 }=9)\ Taking\quad x\quad as\quad \frac { -6+3\sqrt { 19 } }{ 10 } \ { r }^{ 2 }\quad =\quad \frac { 53+6\sqrt { 19 } }{ 5 } \ Clearly\quad a=53,\quad b=6,\quad c=19,\quad d=5\ \dot { .\quad . } \quad a+b+c+d\quad =\quad 83\