the centre of the circle passing through (0,0) and (1,0) and touchng the circle x²+y²=9 ????

fully solved

let circle equation be (x-a)^2+(y-b)^2 =r^2, the circle passes thru origin, thus, (0-a)^2+(0-b)^2=r^2, a^2+b^2=r^2

therefore equation is x^2 -2ax +y^2 -2by=0,it passes thru (1,0) thus, substituting x=1, y=0, we get 1-2a =0, a=1/2

x^2 -2ax +y^2 -2by=0 as before and x^2+y^2=9 touch each other,thus the previous set of two simultaneous equations has only one solution, 9-x-2by=0,9-√(9-y^2)-2by=0, 9-2by=√(9-y^2), y²(4b²+1)-36by+72=0,this quadratic equation in y has only one solution, therefore equating the discriminant to zero and solving the elementary equation in b we get b=±1

thus centre of circle is (0.5,1)or(0.5,-1)

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