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(B) Let the ends of the diameter be $A(2\cos	heta, 2\sin	heta)$ and $B(-2\cos	heta, -2\sin	heta)$.The length of the perpendicular from $(x_1, y_1)

Vedclass · Accepted Answer

$(\sqrt{2}, \sqrt{2}), (-\sqrt{2}, -\sqrt{2})$

Vedclass · Answer

(B) Let the ends of the diameter be $A(2\cos	heta, 2\sin	heta)$ and $B(-2\cos	heta, -2\sin	heta)$.The length of the perpendicular from $(x_1, y_1)$ to $x+y+1=0$ is $d = \frac{|x_1+y_1+1|}{\sqrt{1^2+1^2}} = \frac{|x_1+y_1+1|}{\sqrt{2}}$.Let $d_1$ and $d_2$ be the lengths of the perpendiculars from $A$ and $B$ respectively.$d_1 = \frac{|2\cos	heta+2\sin	heta+1|}{\sqrt{2}}$ and $d_2 = \frac{|-2\cos	heta-2\sin	heta+1|}{\sqrt{2}}$.The product $P = d_1 d_2 = \frac{|(1+2(\cos	heta+\sin	heta))(1-2(\cos	heta+\sin	heta))|}{2} = \frac{|1-4(\cos	heta+\sin	heta)^2|}{2}$.$P = \frac{|1-4(1+2\sin	heta\cos	heta)|}{2} = \frac{|1-4-4\sin(2	heta)|}{2} = \frac{|-3-4\sin(2	heta)|}{2} = \frac{3+4\sin(2	heta)}{2}$.For $P$ to be maximum,$\sin(2	heta) = 1$,which means $2	heta = \frac{\pi}{2}$,so $	heta = \frac{\pi}{4}$.Substituting $	heta = \frac{\pi}{4}$ into the coordinates of $A$ and $B$:$A = (2\cos\frac{\pi}{4}, 2\sin\frac{\pi}{4}) = (2\frac{1}{\sqrt{2}}, 2\frac{1}{\sqrt{2}}) = (\sqrt{2}, \sqrt{2})$.$B = (-2\cos\frac{\pi}{4}, -2\sin\frac{\pi}{4}) = (-\sqrt{2}, -\sqrt{2})$.Thus,the correct option is $B$.

If the product of the lengths of the perpendiculars drawn from the ends of a diameter of the circle $x^2+y^2=4$ onto the line $x+y+1=0$ is maximum,then the two ends of that diameter are

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