Key Concepts and Formulas

• Equation of a Circle: A circle with center $(h, k)$ and radius $r$ has the equation $(x-h)^2 + (y-k)^2 = r^2$.
• Distance from a Point to a Line: The perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
• Intercept Form of a Line: A line with x-intercept $a$ and y-intercept $b$ has the equation $\frac{x}{a} + \frac{y}{b} = 1$.

Step-by-Step Solution

Step 1: Define the Circle's Equation

Since the circle touches both coordinate axes in the first quadrant, its center must be at $(a, a)$ and its radius must be $a$, where $a > 0$. Thus, the equation of the circle is: $(x-a)^2 + (y-a)^2 = a^2$

Step 2: Use the Point P to Find a Relationship

The circle passes through the point $P(\alpha, \beta)$. Substituting the coordinates of $P$ into the circle's equation, we get: $(\alpha - a)^2 + (\beta - a)^2 = a^2$ Expanding and simplifying: $\alpha^2 - 2a\alpha + a^2 + \beta^2 - 2a\beta + a^2 = a^2$ $\alpha^2 + \beta^2 - 2a(\alpha + \beta) + a^2 = 0 \quad (*)$

Step 3: Find the Equation of Line AB

The circle touches the x-axis at $A(a, 0)$ and the y-axis at $B(0, a)$. The equation of the line AB can be found using the intercept form: $\frac{x}{a} + \frac{y}{a} = 1$ Multiplying by $a$, we get: $x + y = a$ Rearranging, we have: $x + y - a = 0$

Step 4: Calculate the Distance PQ

$PQ$ is the perpendicular distance from point $P(\alpha, \beta)$ to the line $x + y - a = 0$. Using the formula for the distance from a point to a line, we have: $PQ = \frac{|\alpha + \beta - a|}{\sqrt{1^2 + 1^2}} = \frac{|\alpha + \beta - a|}{\sqrt{2}}$ We are given that $PQ = 11$, so: $\frac{|\alpha + \beta - a|}{\sqrt{2}} = 11$ $|\alpha + \beta - a| = 11\sqrt{2}$ Since $P$ is above the line $AB$, $\alpha + \beta > a$, so $\alpha + \beta - a > 0$. Therefore, $\alpha + \beta - a = 11\sqrt{2}$ $\alpha + \beta = a + 11\sqrt{2} \quad (**)$

Step 5: Substitute and Solve for 'a'

Substitute equation $(**)$ into equation $(*)$: $\alpha^2 + \beta^2 - 2a(a + 11\sqrt{2}) + a^2 = 0$ $\alpha^2 + \beta^2 - 2a^2 - 22\sqrt{2}a + a^2 = 0$ $\alpha^2 + \beta^2 - a^2 - 22\sqrt{2}a = 0$ Also, square equation $(**)$: $(\alpha + \beta)^2 = (a + 11\sqrt{2})^2$ $\alpha^2 + 2\alpha\beta + \beta^2 = a^2 + 22\sqrt{2}a + 242$ Now substitute $\alpha^2 + \beta^2 = a^2 + 22\sqrt{2}a$ into this equation: $a^2 + 22\sqrt{2}a + 2\alpha\beta = a^2 + 22\sqrt{2}a + 242$ $2\alpha\beta = 242$ $\alpha\beta = 121$

Step 6: Find the value of square root of alpha beta

The question asks for square root of alpha beta $\sqrt{\alpha\beta} = \sqrt{121}$ $\sqrt{\alpha\beta} = 11$

Step 7: Re-examine the problem statement

We made an error in interpreting the question. The problem asks for the value of $\sqrt[5]{\alpha\beta}$ which is $\sqrt[5]{121}$

Step 8: Re-examine the problem statement

We made an error in interpreting the question. The problem asks for the value of the greatest integer less than $\sqrt[5]{\alpha\beta}$ which is 2

Common Mistakes & Tips

• Carefully read the problem statement to understand exactly what needs to be calculated.
• Remember the formula for the distance from a point to a line.
• Be careful with algebraic manipulations and substitutions.

Summary

The problem involves a circle touching both coordinate axes, a point P on the circle, and the perpendicular distance from P to the line connecting the points where the circle touches the axes. By using the equation of the circle, the equation of the line, and the distance formula, we found a relationship between $\alpha$, $\beta$, and $a$. Then we found the $\alpha\beta$, and finally, the greatest integer less than $\sqrt[5]{\alpha\beta}$

Final Answer

The final answer is \boxed{2}.