Question

If the circles described on the line joining the points $(0,1)$ and $(\alpha ,\beta )$ as diameter cuts the $X-$ axis is points whose abscissa are roots of equation $x^2-5x+3=0$ then $(\alpha ,\beta )=$

• A. $(5,3)$
• B. $(3,5)$
• C. $(-5,3)$
• D. $(-5,-3)$

Answer 1

The correct option is A $(5,3)$

$x^2-5x+3=0$

$x_1+x_2=5$

$x_1{x}_2=3$

$x,x_2=3$

$|x_1-x_2|=\sqrt{(x_1+x_2{)}^2-4(x_1{x}_2)}$

$=\sqrt{25-12}$

$|x_1-x_2|=\sqrt{13}$

$O\Rightarrow (\frac{\alpha }{2},\frac{\beta -1}{2})$

$PQ=|x_2-x_1|=\sqrt{13}$

$R\Rightarrow (\frac{x_1+x_2}{2},0)\Rightarrow (\frac{3}{2},0)$

$o$ and $R$ will have same ${}^{\prime }{x}^{\prime }$ coordinates.

$\Rightarrow \frac{\alpha }{2}=\frac{5}{2}$

$\Rightarrow \alpha =5$

$OR=\left(\frac{\beta +1}{2}\right)$ $PQ=\sqrt{13}$ $\Rightarrow RQ=\frac{\sqrt{13}}{2}$

$OQ=r=\frac{{\alpha }^2+(\beta -1{)}^2}{2}$

$O{Q}^2=O{R}^2+R{Q}^2$

$\frac{{\alpha }^2+(\beta -1{)}^2}{4}=\frac{(\beta {)}^2}{4}+\frac{13}{4}$

$\frac{25}{4}=\frac{({\beta }^2+1+2\beta +{\beta }^2-1+2\beta )}{4}+\frac{13}{4}$

$\Rightarrow \beta =3$

$\therefore (\alpha ,\beta )(5,3)$