JEE Advance - Mathematics (1984 - No. 30)
The abscissa of the two points A and B are the roots of the equation $${x^2}\, + \,2ax\, - {b^2} = 0$$ and their ordinates are the roots of the equation $${x^2}\, + \,2px\, - {q^2} = 0$$. Find the equation and the radius of the circle with AB as diameter.
$${x^2}, + ,{y^2} + ,2ax, + 2py, - {b^2}, - {q^2} = 0,,,,sqrt {{a^2}, + ,{p^2} + {b^2}, + ,{q^2}} $$
$${x^2}, + ,{y^2} - ,2ax, - 2py, + {b^2}, + {q^2} = 0,,,,sqrt {{a^2}, + ,{p^2} - {b^2}, - ,{q^2}} $$
$${x^2}, + ,{y^2} + ,2ax, - 2py, - {b^2}, + {q^2} = 0,,,,sqrt {{a^2}, + ,{p^2} + {b^2}, - ,{q^2}} $$
$${x^2}, + ,{y^2} - ,2ax, + 2py, + {b^2}, - {q^2} = 0,,,,sqrt {{a^2}, - ,{p^2} + {b^2}, + ,{q^2}} $$
$${x^2}, + ,{y^2} + ,2ax, + 2py, + {b^2}, + {q^2} = 0,,,,sqrt {{a^2}, - ,{p^2} - {b^2}, - ,{q^2}} $$
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