Two circles in the first quadrant of radii r_ | vedclass

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(D) Since the circles are in the first quadrant and touch both coordinate axes,their centers are $(r, r)$ and their equations are $(x-r)^2 + (y-r)^2 =

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$7$

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(D) Since the circles are in the first quadrant and touch both coordinate axes,their centers are $(r, r)$ and their equations are $(x-r)^2 + (y-r)^2 = r^2$.Expanding this,we get $x^2 + y^2 - 2rx - 2ry + r^2 = 0$.The length of the intercept cut by the line $x+y-2=0$ is given by $2\sqrt{r^2 - d^2} = 2$,where $d$ is the perpendicular distance from the center $(r, r)$ to the line $x+y-2=0$.Thus,$\sqrt{r^2 - d^2} = 1$,which implies $r^2 - d^2 = 1$.The distance $d = \frac{|r+r-2|}{\sqrt{1^2+1^2}} = \frac{|2r-2|}{\sqrt{2}} = \sqrt{2}|r-1|$.Substituting $d^2 = 2(r-1)^2$ into the equation $r^2 - d^2 = 1$,we get $r^2 - 2(r-1)^2 = 1$.$r^2 - 2(r^2 - 2r + 1) = 1$ $\Rightarrow r^2 - 2r^2 + 4r - 2 = 1$ $\Rightarrow -r^2 + 4r - 3 = 0$.So,$r^2 - 4r + 3 = 0$. The roots of this quadratic equation are $r_1$ and $r_2$.By Vieta's formulas,$r_1 + r_2 = 4$ and $r_1 r_2 = 3$.We need to find $r_1^2 + r_2^2 - r_1 r_2 = (r_1 + r_2)^2 - 3r_1 r_2$.Substituting the values,we get $4^2 - 3(3) = 16 - 9 = 7$.

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 $...........$

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