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

Question

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 _ _ _ _ _ _ _

TestHub · Accepted Answer

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,
$\frac{h - α}{1} = \frac{k - β}{1} = \frac{- (α + β - r)}{2}$
$\Rightarrow (h, k) = (α - \frac{α}{2} - \frac{β}{2} + \frac{r}{2}, β - \frac{α}{2} - \frac{β}{2} + \frac{r}{2})$
$\Rightarrow (h, k) = (\frac{α - β + r}{2}, \frac{- α + β + r}{2})$
Now using the distance formula we get,
${(P Q)}^{2} = {(\frac{- α - β + r}{2})}^{2} + {(\frac{- α - β + r}{2})}^{2} = 121$
$\Rightarrow (r - (α + β))^{2} = 242$
$\Rightarrow r^{2} + (α + β)^{2} - 2 r (α + β) = 242$
$\Rightarrow α^{2} + β^{2} + 2 α β + r^{2} - 2 r α - 2 r β = 242$
$\Rightarrow 2 α β = 242$ $\{from equation (1), (α - r)^{2} + (β - r)^{2} = r^{2}\}$
$\Rightarrow α β = 121$

Mathematics - Circle Question with Solution | TestHub

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