Question

If $\alpha ,\beta$, are the roots of the equation $a{x}^2+bx+c=0$ and ${\alpha }^{\prime },{\beta }^{\prime }$ those of $a^{\prime }{x}^2+b^{\prime }x+c^{\prime }=0$ and the circle having $A(\alpha ,{\alpha }^{\prime })$ and $B(\beta ,{\beta }^{\prime })$ as diameter passes through the origin, and the point $(\frac{b}{a},\frac{b^{\prime }}{a^{\prime }})$ then

• A. $a^{\prime }c+a{c}^{\prime }=0$
• B. $a^{\prime }b+a{b}^{\prime }=0$
• C. $b^{\prime }c+b{c}^{\prime }=0$
• D. $a^2{b}^{\prime 2}+a^{\prime 2}{b}^2=0$

Answer 1

Answer: (A), (D)

The correct options are
A $a^{\prime }c+a{c}^{\prime }=0$
D $a^2{b}^{\prime 2}+a^{\prime 2}{b}^2=0$

We have $\alpha +\beta =-b/a,\alpha \beta =c/a,$
${\alpha }^{\prime }+{\beta }^{\prime }=-b^{\prime }/a^{\prime }$ and ${\alpha }^{\prime }{\beta }^{\prime }=c^{\prime }/a^{\prime }$.
Therefore, equation of the circle having $AB$ as diameter is $(x-\alpha )(x-\beta )+(y-{\alpha }^{\prime })(y-{\beta }^{\prime })=0$ $\Rightarrow x^2-(\alpha +\beta )x+\alpha \beta +y^2-({\alpha }^{\prime }+{\beta }^{\prime })y+{\alpha }^{\prime }{\beta }^{\prime }=0$
$\Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}+y^2+\frac{b^{\prime }}{a^{\prime }}y+\frac{c^{\prime }}{a^{\prime }}=0$
$\Rightarrow a{a}^{\prime }(x^2+y^2)+a^{\prime }bx+a{b}^{\prime }y+a^{\prime }c+a{c}^{\prime }=0$
Since it passes througb the origin, $a^{\prime }c+a{c}^{\prime }=0$ and through $(b/a,b^{\prime }/a^{\prime })$
$a{a}^{\prime }(\frac{b^2}{a^2}+\frac{b^{\prime 2}}{a^{\prime 2}})+a^{\prime }b\times \frac{b}{a}+a{b}^{\prime }\times \frac{b^{\prime }}{a^{\prime }}=0$
$a^{\prime 2}{b}^2+a^2{b}^{\prime 2}=0$