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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 linex+y=2. Then r_1^2+r_2^2-r_1 r_2 is equal to

Option: 1

7


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\operatorname{Circle}(x-a)^2+(y-a)^2=a^2\\ \begin{aligned} & 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}

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 \mathrm{a}-2)^2}{2}=1 \Rightarrow 2 a^2-4 a^2+8 a-4=2 \\ & \Rightarrow 2 a^2-8 \mathrm{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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Divya Prakash Singh

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