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 the 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 D None of these

Equation of line AB is $(y-0)=\frac{4-0}{3-1}(x-1)\Rightarrow y=2(x-1)$
Solving line AB with the given circle,
$x^2+4(x-1{)}^2=4\Rightarrow x=0,\frac{8}{5}$ $\Rightarrow y=-2,\frac{6}{5}$
Thus $P=(0,-2)$ and $Q=(\frac{8}{5},\frac{6}{5})$
Also $A=(1,0),B=(3,4)$
Now, $BP=\sqrt{3^2+6^2}=\sqrt{45}=3\sqrt{5},PA=\sqrt{1^2+2^2}=\sqrt{5}\therefore \alpha =3$
and, $BQ=\sqrt{{(3-\frac{8}{5})}^2+{(4-\frac{6}{5})}^2}=\frac{7}{\sqrt{5}}$
$QA=\sqrt{{(1-\frac{8}{5})}^2+{(0-\frac{6}{5})}^2}=\frac{3}{\sqrt{5}}\therefore \beta =\frac{7}{3}$
Now $\alpha +\beta =\frac{16}{3},\alpha .\beta =7$
Hence quadratic equation having roots $\alpha ,\beta$ is given by,
$3x^2-16x+21=0$