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# The abcissa and ordinate of the end points A and B of a focal chord of the parabola y^{2} = 4x are respectively the roots of x^{2} – 3x + a = 0 and y^{2} + 6y + b = 0. The equation of the circle with AB as diameter is.

- x
^{2} + y^{2} – 3x + 6y + 3 = 0
- x
^{2} + y^{2} – 3x + 6y – 3 = 0
- x
^{2} + y^{2} + 3x + 6y – 3 = 0
- x
^{2} + y^{2} – 3x – 6y – 3 = 0

^{2}+ y^{2}– 3x + 6y + 3 = 0^{2}+ y^{2}– 3x + 6y – 3 = 0^{2}+ y^{2}+ 3x + 6y – 3 = 0^{2}+ y^{2}– 3x – 6y – 3 = 0## The correct answer is: x^{2} + y^{2} – 3x + 6y – 3 = 0

### t_{1} t_{2} = – 1 as AB is focal chord

x^{2} – 3x + a = 0 ; x_{1} + x_{2} = 3 & x_{1}x_{2} = a

y^{2} + 6y + b = 0 ; y_{1} + y_{2} = – 6 & y_{1}y_{2} = b

x_{1}x_{2} = . t_{1}^{2} = 1 = a

y_{1}y_{2} = 2t_{1} = – 4 = b

a = 1, b = – 4

equation of circle AB is diameter

x^{2} + y^{2} – 3x + 6y – 3 = 0

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