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Circle

Section: Mathematics

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Question : 41 of 89

Marks: +1, -0

If the product of the lengths of the perpendicular drawn from the ends of a diameter of the circle

x^2+y^2=4

x+y+1=0

is maximum, then the two ends of that diameter are

[AP EAPCET 21 May 2025 S1]

$(-2,0),(2,0)$
$(\sqrt{3}, 1),(-\sqrt{3}, -1)$
$(\sqrt{2}, \sqrt{2}),(-\sqrt{2}, -\sqrt{2})$
$(0,2),(0,-2)$

Solution:

Let the ends of diameter of the given circle

x^2+y^2=4

is

(2\cos\theta, 2\sin\theta)

, then other must be

(-2\cos\theta, -2\sin\theta)

Then, length of perpendicular from the extremities of diameter to the line

x+y+1=0

is

\frac{2(\cos\theta+\sin\theta)+1}{\sqrt{2}}

\times \frac{-2(\cos\theta+\sin\theta)+1}{\sqrt{2}} = P(\text{say})

\Rightarrow P = \frac{1-4(\cos\theta+\sin\theta)^2}{2}

= \frac{1-4(1+\sin 2\theta)}{2}

\Rightarrow P = \frac{-3-4\sin 2\theta}{2}

\Rightarrow \frac{dP}{d\theta} = \frac{1}{2}(-4)\cos 2\theta \cdot 2 = 0

\Rightarrow \cos 2\theta = 0 \Rightarrow 2\theta = \frac{\pi}{2}

\theta = \frac{\pi}{4}

Hence, the points are

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

and

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

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