JEE MAIN - Mathematics (2023 - 6th April Morning Shift - No. 19)
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 $$A B$$. 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 $$\alpha \beta$$ is
___________.
Answer
121
Explanation
Let equation of circle is $(x-a)^2+(y-a)^2=a^2$
Since, (i) passes through $P(\alpha, \beta)$
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$\therefore (\alpha-a)^2+(\beta-a)^2=a^2$
$$ \begin{array}{lr} &\Rightarrow \alpha^2+\beta^2-2 \alpha a-2 \beta a+a^2=0 .........(i) \end{array} $$
Equation of $A B=\frac{x}{a}+\frac{y}{a}=1$
$$\Rightarrow x+y=a$$ .........(ii)
Let $Q(p, q)$ be the foot of the perpendicular from $P$ to line (iii)
$$ \begin{aligned} & \therefore \frac{p-\alpha}{1}=\frac{q-\beta}{1}=\frac{-(\alpha+\beta-\alpha)}{(1)^2+(1)^2} \\\\ & \Rightarrow \frac{p-\alpha}{1}=\frac{q-\beta}{1}=\frac{-(\alpha+\beta-a)}{2} \\\\ & \Rightarrow p-\alpha=\frac{-(\alpha+\beta-a)}{2} \\\\ & \text { and } q-\beta=\frac{-(\alpha+\beta-a)}{2} \end{aligned} $$
Now,
$$ \begin{aligned} & P Q^2 =(p-\alpha)^2+(q-\beta)^2 \\\\ & =\frac{1}{4}(\alpha+\beta-a)^2+\frac{1}{4}(\alpha+\beta-a)^2 \end{aligned} $$
$$ \Rightarrow (11)^2=\frac{1}{2}(\alpha+\beta-\alpha)^2 $$
$$ \begin{array}{ll} &\Rightarrow \alpha^2+\beta^2+a^2+2 \alpha \beta-2 \beta a-2 \alpha a=242 \\\\ &\Rightarrow 2 \alpha \beta=242 \text { [Using Eq. (ii)] }\\\\ &\Rightarrow \alpha \beta=121 \end{array} $$
Since, (i) passes through $P(\alpha, \beta)$
$\therefore (\alpha-a)^2+(\beta-a)^2=a^2$
$$ \begin{array}{lr} &\Rightarrow \alpha^2+\beta^2-2 \alpha a-2 \beta a+a^2=0 .........(i) \end{array} $$
Equation of $A B=\frac{x}{a}+\frac{y}{a}=1$
$$\Rightarrow x+y=a$$ .........(ii)
Let $Q(p, q)$ be the foot of the perpendicular from $P$ to line (iii)
$$ \begin{aligned} & \therefore \frac{p-\alpha}{1}=\frac{q-\beta}{1}=\frac{-(\alpha+\beta-\alpha)}{(1)^2+(1)^2} \\\\ & \Rightarrow \frac{p-\alpha}{1}=\frac{q-\beta}{1}=\frac{-(\alpha+\beta-a)}{2} \\\\ & \Rightarrow p-\alpha=\frac{-(\alpha+\beta-a)}{2} \\\\ & \text { and } q-\beta=\frac{-(\alpha+\beta-a)}{2} \end{aligned} $$
Now,
$$ \begin{aligned} & P Q^2 =(p-\alpha)^2+(q-\beta)^2 \\\\ & =\frac{1}{4}(\alpha+\beta-a)^2+\frac{1}{4}(\alpha+\beta-a)^2 \end{aligned} $$
$$ \Rightarrow (11)^2=\frac{1}{2}(\alpha+\beta-\alpha)^2 $$
$$ \begin{array}{ll} &\Rightarrow \alpha^2+\beta^2+a^2+2 \alpha \beta-2 \beta a-2 \alpha a=242 \\\\ &\Rightarrow 2 \alpha \beta=242 \text { [Using Eq. (ii)] }\\\\ &\Rightarrow \alpha \beta=121 \end{array} $$
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