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  • The circle  <img alt="\mathrm{x^2+y^2+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2&plus;2%20g%20x&plus;2%20f%20y&plu

The circle  \mathrm{x^2+y^2+2 g x+2 f y+c=0} cuts off an intercept on the straight line \mathrm{lx+my}=l which subtends an angle of 45° at the centre of the circle. If \mathrm{\left(l^2+m^2\right)\left(g^2+f-c\right)=(k-2 \sqrt{2})(l g+m f+l)^2} , then k=

 

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

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\mathrm{(-g,-f)} is the centre and \mathrm{\sqrt{g^2+f^2-c}} is the radius of the given circle.The length of perpendicular CM on intercept AB from centre C is

\mathrm{\mathrm{CM}=\left|\frac{\mathrm{l}(-\mathrm{g})+\mathrm{m}(-\mathrm{f})-1}{\sqrt{1^2+\mathrm{m}^2}}\right|}

Since, \mathrm{\angle \mathrm{ACB}=45^{\circ}, \angle \mathrm{ACM}=22 \frac{1^{\circ}}{2}}

From right triangle ACM,


\mathrm{\begin{aligned} & \cos 22^{\frac{1^{\circ}}{2}}=\frac{C M}{A C} \\ & \frac{\sqrt{2+\sqrt{2}}}{2}=\left|\frac{\lg +\mathrm{mf}+1}{\sqrt{1^2+\mathrm{m}^2} \sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}}\right| \end{aligned}}

\mathrm{\Rightarrow\left(l^2+\mathrm{m}^2\right)\left(\mathrm{g}^2+\mathrm{f}^2=1\right)=(4-2 \sqrt{2})(l \mathrm{~g}+\mathrm{mf}+1)^2}

 

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Sayak

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