The angle between the pair of straight lines | vedclass

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(A) The equation of the circle is $x^2+y^2=4$ and the line is $y=3x+c$,which can be written as $\frac{y-3x}{c}=1$. To find the equation of the pair of

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$20$

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(A) The equation of the circle is $x^2+y^2=4$ and the line is $y=3x+c$,which can be written as $\frac{y-3x}{c}=1$. To find the equation of the pair of lines joining the origin to the intersection points,we homogenize the circle equation: $x^2+y^2=4(1)^2$ $x^2+y^2=4\left(\frac{y-3x}{c}\right)^2$ $c^2(x^2+y^2)=4(y^2+9x^2-6xy)$ $c^2x^2+c^2y^2=4y^2+36x^2-24xy$ $(c^2-36)x^2+24xy+(c^2-4)y^2=0$ Since the lines are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(c^2-36)+(c^2-4)=0$ $2c^2-40=0$ $2c^2=40$ $c^2=20$

The angle between the pair of straight lines formed by joining the points of intersection of $x^2+y^2=4$ and $y=3x+c$ to the origin is a right angle. Then $c^2$ is equal to

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