there are two ways of solving it
Ist
applying two pnt formula (dist b/n x1,y1 and x2,y2 is sqrt[(x1-x2)^2+(y1-y2)^2]............1
and pythagoras theorem i.e a^2+b^2=htpotenuous^2..........2
from1 we get s^2+t^2=4...........3
and from 2 we get t=s Sqrt(3).........4
putting 4 in 3 we get s=1
another way is the fig is symmetrical about y axis i.e mirror image so the values get interchanged
i.e x,y will become y,-x
so -rut3,1 will become 1,rut 3
hence s=1
Though your answer is correct, there is some mistake in your calculation as t is not equal to s. And I feel we cannot use the second method you suggested as we do not know if the axis of symmetry coincides with the y axis. The answer seems to suppose that but we cannot jump to that.
Solution
From distance formula we can find distance between (root3, 1) and (0,0) which turns out to be 2 which is the radius
Hence s^2 + t^2 = 4 --1
Now using distance formula between points (root3,1) and (s,t) which will be equal to the hypotenuse of the triangle which is (1,1,root2) triplet hence hyp = 2root2
On equating distance formula and hyp, we obtain
s = t/root3
Substitute in 1 and we get
s= 1, t=root3
which is the answer, s= 1
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