Question

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

• A. $x^2+ax+b=0$
• B. $x^2-ax+b=0$
• C. $x^2+ax-b=0$
• D. $x^2-ax-b=0$

Answer 1

The correct option is B $x^2-ax+b=0$

Coordinates of center $(\frac{a+0}{2},\frac{b+1}{2})=(\frac{a}{2},\frac{b+1}{2})$

Radius $=\frac{1}{2}\sqrt{a^2+{(b-1)}^2}$

Equation of circle

${(x-\frac{a}{2})}^2+{(y-\frac{(b+1)}{2})}^2=\frac{1}{4}(a^2+{(b-1)}^2)$

The circle touches/cuts X-axis at

${(x-\frac{a}{2})}^2+{\left(\frac{b+1}{2}\right)}^2=\frac{1}{4}(a^2+{(b-1)}^2){(x-\frac{a}{2})}^2=\frac{a^2}{4}+[\frac{(b-1{)}^2}{2}-\frac{(b+1{)}^2}{2}]{(x-\frac{a}{2})}^2=\frac{a^2}{4}+\frac{1}{4}\left(2b\right)(-2)=\frac{a^2}{4}-b{(x-\frac{a}{2})}^2=\frac{a^2-4b}{4}{x}^2-ax+\frac{a^2}{4}=\frac{a^2}{4}-b{x}^2-ax+b=0$