JEE MAIN - Mathematics (2023 - 12th April Morning Shift - No. 15)
Two circles in the first quadrant of radii $$r_{1}$$ and $$r_{2}$$ touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line $$x+y=2$$. Then $$r_{1}^{2}+r_{2}^{2}-r_{1} r_{2}$$ is equal to ___________.
Answer
7
Explanation
$$
\begin{aligned}
& \text { Circle }(x-a)^2+(y-a)^2=a^2 \\\\
& x^2+y^2-2 a x-2 a y+a^2=0 \\\\
& \text { intercept }=2 \\\\
& \Rightarrow 2 \sqrt{a^2-d^2}=2
\end{aligned}
$$
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Where $\mathrm{d}=$ perpendicular distance of centre from line $x+y=2$
$$ \begin{aligned} & \Rightarrow 2 \sqrt{a^2-\left(\frac{a+a-2}{\sqrt{2}}\right)^2}=2 \\\\ & \Rightarrow a^2-\frac{(2 a-2)^2}{2}=1 \Rightarrow 2 a^2-4 a^2+8 a-4=2 \\\\ & \Rightarrow 2 a^2-8 a+6=0 \Rightarrow a^2-4 a+3=0 \\\\ & \therefore r_1+r_2=4 \text { and } r_1 r_2=3 \\\\ & \therefore r_1^2+r_2^2-r_1 r_2=\left(r_1+r_2\right)^2-3 r_1 r_2 \\\\ & =16-9=7 \end{aligned} $$
Where $\mathrm{d}=$ perpendicular distance of centre from line $x+y=2$
$$ \begin{aligned} & \Rightarrow 2 \sqrt{a^2-\left(\frac{a+a-2}{\sqrt{2}}\right)^2}=2 \\\\ & \Rightarrow a^2-\frac{(2 a-2)^2}{2}=1 \Rightarrow 2 a^2-4 a^2+8 a-4=2 \\\\ & \Rightarrow 2 a^2-8 a+6=0 \Rightarrow a^2-4 a+3=0 \\\\ & \therefore r_1+r_2=4 \text { and } r_1 r_2=3 \\\\ & \therefore r_1^2+r_2^2-r_1 r_2=\left(r_1+r_2\right)^2-3 r_1 r_2 \\\\ & =16-9=7 \end{aligned} $$
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