The circle described on the line joining the points (0,1),(a, b) as diameter cuts the x-axis in points whose abscissa are roots of the equation:

Question

The circle described on the line joining the points (0,1),(a, b) as diameter cuts the x-axis in points whose abscissa are roots of the equation: (A) x2+a x+b=0 (B) x2-a x+b=0 (C) x2+a x-b=0 (D) x2-a x-b=0

Tardigrade Admin · Accepted Answer

Equation of circle (x-0)(x-a)+(y-1)(y-b)=0 it cuts x-axis put y=0 ⇒ x2-a x+b=0

Q. The circle described on the line joining the points $(0, 1), (a, b)$ as diameter cuts the $x$ -axis in points whose abscissa are roots of the equation:

$x^{2} + a x + b = 0$

$x^{2} - a x + b = 0$

$x^{2} + a x - b = 0$

$x^{2} - a x - b = 0$

Equation of circle $(x - 0) (x - a) + (y - 1) (y - b) = 0$
it cuts $x$ -axis put $y = 0 \Rightarrow x^{2} - a x + b = 0$

Q. The circle described on the line joining the points (0,1),(a,b) as diameter cuts the x-axis in points whose abscissa are roots of the equation:

x2+ax+b=0

x2−ax+b=0

x2+ax−b=0

x2−ax−b=0