The pair of straight lines joining the origin to the common points $x^{2} + y^{2} = 4$ and $y = 3 x + c$ are perpendicular if $c^{2}$ =

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Solution

The correct option is A 20
$x^{2} + y^{2} = 4$ , $y = 3 x + c$

using homogenization of curves

$\frac{y - 3 x}{c} = 1$

$x^{2} + y^{2} - 4 {(\frac{y - 3 x}{c})}^{2} = 0$

Now, if pair is $⊥$ , then

Coefficient of $x^{2}$ + Coefficient of $y^{2} = 0$
$a + b = 0$

Here
$a = 1 - \frac{36}{c^{2}}$ , $b = 1 - \frac{4}{c^{2}}$

$a + b = 0$

$2 - \frac{40}{c^{2}} = 0$

$c^{2} = 20$

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The pair of straight lines joining the origin to the common points x2+y2=4 and y=3x+c are perpendicular if c2 =

The correct option is A 20x2+y2=4, y=3x+cusing homogenization of curvesy−3xc=1x2+y2−4(y−3xc)2=0Now, if pair is ⊥, thenCoefficient of x2 + Coefficient of y2=0a+b=0Herea=1−36c2, b=1−4c2a+b=02−40c2=0c2=20

The pair of straight lines joining the origin to the common points $x^{2} + y^{2} = 4$ and $y = 3 x + c$ are perpendicular if $c^{2}$ =