If it is possible to draw a triangle which circumscribes the circle `(x-(a-2b))^2+(y-(a+b))^2=1` and is inscribed by `x^2+y^2-2x-4y+1=0` thenA. `beta = -(1)/(3)`B. `beta =(2)/(3)`C. `alpha=(5)/(3)`D. `alpha = - (5)/(2)`

If it is possible to draw a triangle which circumscribes the circle `(

1.	If it is possible to draw a triangle which circumscribes the circle `(x-(a-2b))^2+(y-(a+b))^2=1` and is inscribed by `x^2+y^2-2x-4y+1=0` thenA. `beta = -(1)/(3)`B. `beta =(2)/(3)`C. `alpha=(5)/(3)`D. `alpha = - (5)/(2)`
Answer» Correct Answer - 3 For such triangle, first circle is the incircle and other one is the circumcircle. Also, the radius of first circle is half of the radius of the second circle. So, triangle is equilateral Thus, incentre and circumcentre coincide. `implies alpha-2beta=1` and `alpha+beta=2` `implies (alpha,beta)-=(5)/(3),(1)/(3)`

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