A circle passing through the point P α , β in the first quadrant touches the two coordinate axes at the points A and B . The point P is above the line A B . The point Q on the line segment A B is the foot of perpendicular from P on A B . If P Q is equal to 11 units, then the value of α β is _ _ _ _ _ _ _

Step-by-step solution

Let the equation of circle touching both the coordinate axis be, ( x - r ) 2 + ( y - r ) 2 = r 2 which passes through α , β So, ( α - r ) 2 + ( β - r ) 2 = r 2   . . . . . . . 1 Also equation of line A B :   x + y = r Let Q h , k , Now using the formula of foot of perpendicular, h - α 1 = k - β 1 = - ( α + β - r ) 2 ⇒ ( h , k ) = α - α 2 - β 2 + r 2 , β - α 2 - β 2 + r 2 ⇒ h , k = α - β + r 2 , - α + β +