The circle x2 - y2 - 4 cuts the line joining the points A1- 0 and B3- 4 in two points P and Q- Let BP-PA - - and BQ-QA - - Then - and - are roots of the quadratic equation

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Length of Intercept Made by a Circle on a Straight Line

The circle x2...

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 $\frac{B P}{P A} = α$ and

$\frac{B Q}{Q A} = β$ . Then $α a n d β$ are roots of the quadratic equation

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

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$3 x^{2} + 2 x + 21 = 0$

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

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None of these

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Solution

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

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

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The circle x2+y2=4 cuts the line joining the points A(1, 0) and B(3, 4) in two points P and Q. Let BPPA=α and BQQA=β. Then α and β are roots of the quadratic equation

The correct option is A 3x2+2x−21=0 x2+y2=4 A(1,0) B(3,4) Equation of line joining AB is y=2x−2 Substituting this in circle equation we have x2+(2x−2)2=4 x=0,85 y=−2,65 P=(85,65) Q=(0,−2) We get α=73,β=3 Quadratic equation which have α,β as roots is x2−(73+3)x+7=0 3x2−16x+21=0

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 $\frac{B P}{P A} = α$ and

$\frac{B Q}{Q A} = β$ . Then $α a n d β$ are roots of the quadratic equation