Consider a circle $x - \alpha ^ { 2 } + y - \beta ^ { 2 } = 50$, where $\alpha , \beta > 0$. If the circle touches the line $y + x = 0$ at the point P , whose distance from the origin is $4 \sqrt { 2 }$, then $( \alpha + \beta ) ^ { 2 }$ is equal to $\_\_\_\_$ .

Let the circles $C _ { 1 } : ( x - \alpha ) ^ { 2 } + ( y - \beta ) ^ { 2 } = r _ { 1 } ^ { 2 }$ and $C _ { 2 } : ( x - 8 ) ^ { 2 } + \left( y - \frac { 15 } { 2 } \right) ^ { 2 } = r _ { 2 } ^ { 2 }$ touch each other externally at the point $( 6,6 )$. If the point $( 6,6 )$ divides the line segment joining the centres of the circles $C _ { 1 }$ and $C _ { 2 }$ internally in the ratio $2 : 1$, then $( \alpha + \beta ) + 4 \left( r _ { 1 } ^ { 2 } + r _ { 2 } ^ { 2 } \right)$ equals
(1) 125
(2) 130
(3) 110
(4) 145

A circle $C$ of radius 2 lies in the second quadrant and touches both the coordinate axes. Let $r$ be the radius of a circle that has centre at the point $( 2,5 )$ and intersects the circle $C$ at exactly two points. If the set of all possible values of r is the interval $( \alpha , \beta )$, then $3 \beta - 2 \alpha$ is equal to:
(1) 10
(2) 15
(3) 12
(4) 14

We have a triangle ABC such that
$$\mathrm { AB } = 9 , \quad \mathrm { BC } = 12 , \quad \angle \mathrm { ABC } = 90 ^ { \circ } .$$
There are also two circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ with radii of length $2r$ and $r$, respectively. The two circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ are tangential to each other. Further, $\mathrm { O } _ { 1 }$ is tangent to the two sides AB and AC, and $\mathrm { O } _ { 2 }$ is tangent to the two sides CA and CB. We are to find the value of $r$.
First, let D and E denote the points at which the segment AC is tangent to the circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ respectively, and set $\alpha = \angle \mathrm { O } _ { 1 } \mathrm { AC }$. Then, since $\tan 2 \alpha = \frac { \square \mathbf { A } } { \square }$, we have $\tan \alpha = \frac { \mathbf { C } } { \mathbf { D } }$ using the double-angle formula. Thus, we obtain $\mathrm { AD } = \mathbf { E }$.
Next, set $\beta = \angle \mathrm { O } _ { 2 } \mathrm { CA }$. Since $\alpha + \beta = \mathbf { F G } ^ { \circ }$, we have $\tan \beta = \frac { \mathbf { H } } { \square \mathbf{I} }$ using the addition theorem. Thus, we obtain $\mathrm { CE } = \square r$.
Moreover, it follows that $\mathrm { AC } = \mathbf { K L }$ and $\mathrm { DE } = \mathbf { M } \sqrt { \mathbf { N } } r$.
Finally we obtain
$$r = \frac { \mathbf { O P } ( \mathbf { Q } - \mathbf { R } \sqrt { \mathbf { S } } ) } { 41 }$$

On a coordinate plane, there is a circle with center $A(a, b)$ that is tangent to both coordinate axes. There is also a point $P(c, c)$, where $a > c > 0$, and it is known that $\overline{PA} = a + c$. Select the correct options.
(1) $a = b$ (2) Point $P$ is on the line $x + y = 0$ (3) Point $P$ is inside the circle (4) $\frac{a + c}{b - c} = \sqrt{2}$ (5) $\frac{a}{c} = 2 + 3\sqrt{2}$

Below, two concentric circles with center $O$ and a circle with center $M$ tangent to both circles are given.
The radius of the small circle with center $O$ is 4 units less than the radius of the large circle with center $O$, and 1 unit more than the radius of the circle with center $M$.
Accordingly, what is the area of the shaded region in square units?
A) $28 \pi$ B) $32 \pi$ C) $36 \pi$ D) $39 \pi$ E) $45 \pi$