Question

The circle $x^2+y^2=4$ cuts the line joining the points $A\left(1,0\right)$ and $B\left(3,4\right)$ in two points $P$ and $Q$ . Let, $\frac{PB}{PA}=\alpha$ and $\frac{BQ}{AQ}=\beta$ , then which of the following quadratic equation has either $\alpha$ or $\beta$ as one of their root

• A. $x^2+2x+7=0$
• B. $3x^2+2x-21=0$
• C. $2x^2+3x-27=0$
• D. none of these

Answer 1

Answer: (B)

$\because \frac{BP}{PA}=\alpha$
$BP:PA=\alpha :1$
$\therefore$ Coordinates of $P$ is $\left(\frac{3+\alpha }{1+\alpha },\frac{4}{\alpha +1}\right)$
$P$ lie on $x^2+y^2=4$
$\Rightarrow (\alpha +3{)}^2+16=4(\alpha +1{)}^2$
$\Rightarrow 3{\left(\alpha \right)}^2+2\alpha -21=0\dots \left(i\right)$
and $\frac{BQ}{QA}=\frac{\beta }{1}$
$\Rightarrow BQ:QA=\beta :1$
$\Rightarrow \frac{BQ}{QA}-1=\beta -1$
$\Rightarrow \frac{AB}{QA}=\frac{(\beta -1)}{1}$
$\Rightarrow AB:QA=(\beta -1):1$
$\therefore$ Coordinates of $Q$ is $\left(\frac{\beta -3}{\beta -1},-\frac{4}{\beta -1}\right)$
$Q$ lie on $x^2+y^2=4$
$\therefore (\beta -3{)}^2+16=4(\beta -1{)}^2$
$\Rightarrow 3{\beta }^2-21\beta -21=0$
Hence, $\alpha$ is a roots of $3x^2+2x-21=0$ and
$\beta$ is a root of $3x^2-2x-21=0$ .