Question

A circle passing through the point P\(( \alpha , \beta )\) in the first quadrant touches the two coordinate axes at the points A and B. The point P is above the line AB. The point Q on the line segment AB is the foot of perpendicular from P on AB. If PQ is equal to 11 units, then value of \(( \alpha \beta )\)  is ___

Answer 1

Correct Answer: 121

C : (x - r)² + (y-r)² = r²; 
α² + β² - 2r(α + β) + r² = 0 

α² + β² - 2r(11\(\sqrt{2}\) + r) + r2 = 0
α² + β² - 22\(\sqrt{2}\)r - r² = 0
PF = \(\frac{α+β-r}{\sqrt{2}}\) = 11 
α + B = 11\(\sqrt{2}\) + r 
α² + β² + 2αβ = 242 + r² + 22r\(\sqrt{2}\) 
αβ = 121

So, the answer is 121.

Answer 2

Given Equation of Circle: 

\[ (x - a)^2 + (y - a)^2 = a^2 \]

Point P(\(\alpha, \beta\)):

\[ (\alpha - a)^2 + (\beta - a)^2 = a^2 \]

Simplification:

\[ \alpha^2 + \beta^2 - 2a\alpha - 2a\beta = 0 \]

Equation of Line AB:

\[ x + y = a \]

Foot of Perpendicular Q(\(\alpha', \beta'\)):

\[ \alpha' - \alpha = \beta' - \beta = -\frac{(\alpha + \beta - a)}{2} \]

Distance \( PQ^2 \):

\[ PQ^2 = \frac{(\alpha + \beta - a)^2}{2} = 121 \]

Solving for \(\alpha\beta\):

\[ 242 = \alpha^2 + \beta^2 - 2a\alpha - 2a\beta + a^2 + 2\alpha\beta \] \[ 242 = 2\alpha\beta \implies \alpha\beta = 121 \]