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. $3x^2-16x+21=0$
• B. $3x^2+2x+21=0$
• C. $2x^2+3x-21=0$
• D. $2x^2+3x+21=0$

Answer 1

The correct option is A $3x^2-16x+21=0$

Given circle $S=x^2+y^2=4^2$

$A(1,0),B(3,4)$

Line joining $AB$ is given by

$\frac{y-0}{x-1}=\frac{4–0}{3–1}⟹2x–y=2$

$⟹y=2x-2$

Substituting in $S$

$x^2+4(x–1{)}^2=4$

$⟹5x^2=8x⟹x=0,\frac{8}{5}andy=-2,\frac{6}{5}$

$P=(0,-2),Q=(\frac{8}{5},\frac{6}{5})$

$\alpha =\frac{BP}{PA}=\frac{\sqrt{3^2+6^2}}{\sqrt{1^2+2^2}}=\frac{\sqrt{45}}{\sqrt{5}}=3$

$\beta =\frac{BQ}{QA}=\frac{\sqrt{\frac{7^2}{5^2}+\frac{{14}^2}{5^2}}}{\sqrt{\frac{3^2}{5^2}+\frac{6^2}{5^2}}}$

Equation having $\alpha ,\beta$ as roots is given

$(x-\alpha )(x-\beta )=0$

$⟹(x–3)(x-\frac{7}{3})=0$

$⟹3x^2-16x+21=0$ is the equation.