Question

The circle $x^2+y^2=4$ cuts the line joining the points A(1, 0) and B(3, 4) in two points P and Q. Let $\frac{BP}{PA}=\alpha$ and

$\frac{BQ}{QA}=\beta$. Then $\alpha and\beta$ are roots of the quadratic equation

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

Answer 1

The correct option is A

$3x^2+2x-21=0$

$x^2+y^2=4$
$A(1,0)$
$B(3,4)$
Equation of line joining AB is
$y=2x-2$
Substituting this in circle equation we have
$x^2+(2x-2{)}^2=4$
$x=0,\frac{8}{5}$
$y=-2,\frac{6}{5}$
$P=(\frac{8}{5},\frac{6}{5})$
$Q=(0,-2)$
We get $\alpha =\frac{7}{3},\beta =3$
Quadratic equation which have $\alpha ,\beta$ as roots is
$x^2-(\frac{7}{3}+3)x+7=0$
$3x^2-16x+21=0$