Test Index

Circles Part 1

Section: Mathematics

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Question : 88 of 100

Marks: +1, -0

If the pair of lines

a x^2+2(a+b) x y+b y^2=0

lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then

$3a^2 - 10ab + 3b^2 = 0$
$3a^2 - 2ab + 3b^2 = 0$
$3a^2 + 10ab + 3b^2 = 0$
$3a^2 + 2ab + 3b^2 = 0$

Solution:

As per question area of one sector

=3

area of another sector

\Rightarrow

at center by one sector

=3 \times

angle at center by another sector

Let one angle be

\theta

then other

=30

Clearly

\theta+3\theta=180 \Rightarrow \theta=45^{\circ}

\therefore

Angle between the diameters represented by combined equation

a x^2+2(a+b x y)+b y^2=0

is

45^{\circ}

\therefore

Using

\tan \theta = \frac{2 \sqrt{h^2 - a b}}{a+b}

we get

\tan 45^{\circ} = \frac{2 \sqrt{(a+b)^2 - a b}}{a+b}

\Rightarrow 1 = \frac{2 \sqrt{a^2 + b^2 + a b}}{a+b}

\Rightarrow (a+b)^2 = 4 (a^2 + b^2 + a b)

\Rightarrow a^2 + b^2 + 2ab = 4a^2 + 4b^2 + 4ab

\Rightarrow 3a^2 + 3b^2 + 2ab = 0

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