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QJEEM23Ma6356: JEE Main 2023 Mathematics - Class 11 JEEM23Ma6356 Solution: Two circles in the first quadrant of radii and touch the…

Two circles in the first quadrant of radii r1r_{1} and r2r_{2} touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line x+y=2x+y=2 . Then r12+r22−r1r2r_{1}^{2}+r_{2}^{2}-r_{1} r_{2} is equal to ___________.

Step-by-Step Verified Solution

amp; Circle (x−a)2+(y−a)2=a2amp;x2+y2−2ax−2ay+a2=0amp; intercept =2amp;⇒2a2−d2=2\begin{aligned}& \text { Circle }(x-a)^2+(y-a)^2=a^2 \\\\& x^2+y^2-2 a x-2 a y+a^2=0 \\\\& \text { intercept }=2 \\\\& \Rightarrow 2 \sqrt{a^2-d^2}=2\end{aligned}

JEE Main 2023 (Online) 12th April Morning Shift Mathematics - Circle Question 34 English Explanation

Where d=\mathrm{d}= perpendicular distance of centre from line x+y=2x+y=2

amp;⇒2a2−(a+a−22)2=2amp;⇒a2−(2a−2)22=1⇒2a2−4a2+8a−4=2amp;⇒2a2−8a+6=0⇒a2−4a+3=0amp;∴r1+r2=4 and r1r2=3amp;∴r12+r22−r1r2=(r1+r2)2−3r1r2amp;=16−9=7\begin{aligned}& \Rightarrow 2 \sqrt{a^2-\left(\frac{a+a-2}{\sqrt{2}}\right)^2}=2 \\\\& \Rightarrow a^2-\frac{(2 a-2)^2}{2}=1 \Rightarrow 2 a^2-4 a^2+8 a-4=2 \\\\& \Rightarrow 2 a^2-8 a+6=0 \Rightarrow a^2-4 a+3=0 \\\\& \therefore r_1+r_2=4 \text { and } r_1 r_2=3 \\\\& \therefore r_1^2+r_2^2-r_1 r_2=\left(r_1+r_2\right)^2-3 r_1 r_2 \\\\& =16-9=7\end{aligned}

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