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QJEEM23Ma56ec: JEE Main 2023 Mathematics - Class 11 JEEM23Ma56ec Solution: A circle passing through the point in the first quadrant…

A circle passing through the point P(α,β)P(\alpha, \beta) in the first quadrant touches the two coordinate axes at the points AA and BB . The point PP is above the line ABA B . The point QQ on the line segment ABA B is the foot of perpendicular from PP on ABA B . If PQP Q is equal to 11 units, then the value of αβ\alpha \beta is ___________.

Step-by-Step Verified Solution

Let equation of circle is (x−a)2+(y−a)2=a2(x-a)^2+(y-a)^2=a^2

Since, (i) passes through P(α,β)P(\alpha, \beta)

JEE Main 2023 (Online) 6th April Morning Shift Mathematics - Circle Question 28 English Explanation

∴(α−a)2+(β−a)2=a2\therefore (\alpha-a)^2+(\beta-a)^2=a^2

amp;⇒α2+β2−2αa−2βa+a2=0.........(i)\begin{array}{lr}&\Rightarrow \alpha^2+\beta^2-2 \alpha a-2 \beta a+a^2=0 .........(i)\end{array}

Equation of AB=xa+ya=1A B=\frac{x}{a}+\frac{y}{a}=1

⇒x+y=a\Rightarrow x+y=a .........(ii)

Let Q(p,q)Q(p, q) be the foot of the perpendicular from PP to line (iii)

amp;∴p−α1=q−β1=−(α+β−α)(1)2+(1)2amp;⇒p−α1=q−β1=−(α+β−a)2amp;⇒p−α=−(α+β−a)2amp; and q−β=−(α+β−a)2\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,

amp;PQ2=(p−α)2+(q−β)2amp;=14(α+β−a)2+14(α+β−a)2\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}

⇒(11)2=12(α+β−α)2\Rightarrow (11)^2=\frac{1}{2}(\alpha+\beta-\alpha)^2

amp;⇒α2+β2+a2+2αβ−2βa−2αa=242amp;⇒2αβ=242 [Using Eq. (ii)] amp;⇒αβ=121\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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