A circle touches the parabola y^2=4ax at P. It also possess through the focus S of the parabola and intersects is axis at Q. If angle SPQ=90 degree, find equation of circle?
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As SQ is a diameter of the circle
Let C be it's centre. Then coordinates of C are (a+α/2 ,0)
Radius of the circle is(a-α)^2/2 = (a-α)
From The general equation of the circle
[(2x-a-α)/2]^2+y^2= [(a-α)/2}^2
solving this we get
x^2+y^2-x(a+α/2)+2aα=0
Comparing it with the diametrical form of this circle i.e. x^2-x(a+α)+y^2=0
Clearly 2aα=0 and α=α/2
Thus α=0
Hence Equation of the circle is
x^2-ax+y^2=0
As SQ is a diameter of the circle
Let C be it's centre. Then coordinates of C are (a+α/2 ,0)
Radius of the circle is(a-α)^2/2 = (a-α)
From The general equation of the circle
[(2x-a-α)/2]^2+y^2= [(a-α)/2}^2
solving this we get
x^2+y^2-x(a+α/2)+2aα=0
Comparing it with the diametrical form of this circle i.e. x^2-x(a+α)+y^2=0
Clearly 2aα=0 and α=α/2
Thus α=0
Hence Equation of the circle is
x^2-ax+y^2=0
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