CIRCLE : the circle described on line joining points (0,1),(a,b)as diameter cuts thex axis in points whose abscissae  are roots of equation?

u know the diameter end pt form of equation of a circle..

i.e.

now putting y=0..u get

so rq. equation is ..

x^2 - ax + b = 0..

the centre of circle is (a/2,(b+1)/2)

the radius is [(a²+(b-1)²)^1/2]/2 and lets say this as r

clearly equation is (x-a/2)²+(y-b/2-1/2)²=r²

lets say the points where the circle cuts the x axis as (m1,0) and (m2,0)

substitute these in the circle equation,

(m1-a)²+(b/2+1/2)²=r²

since even m2 in place of m1 satisfies this equation we can denote m1 and m2 by common variable m

and say that m1 and m2 are roots of this quadratic in m

hence our quadratic is (m-a)²+(b/2+1/2)²=r²

DECODER                     why did u put y=0

because on x axis y co ordinate is zero