Test Index

Circles Part 2

Section: Mathematics

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Question : 24 of 87

Marks: +1, -0

Let circle

C

be the image of

x^2+y^2-2 x+4 y-4=0

2 x-3 y+5=0

A

be the point on

C

O A

x

-axis and A lies on the righthand side of the centre

O

C

B(\alpha, \beta)

\beta < 4

C

such that the length of the

\text{arc} AB

\left(\frac{1}{6}\right)^{\text{th}}

of the perimeter of

C

\beta - \sqrt{3} \alpha

[24 Jan 2025 Shift 1]

4
3
$4-\sqrt{3}$
$3+\sqrt{3}$

Solution: 👈: Video Solution

\;x^2+y^2-2 x+4 y-4=0

\;\;\text{ Centre }\;=(1,-2)

\;\;\text{ Radius }\;=\sqrt{1+4+4}=3

2 x-3 y+5=0

\;\;\frac{h-1}{2}=\;\frac{k+2}{-3}=\;\frac{-2(2+6+5)}{2^2+(-3)^2}

\;\Rightarrow\;\frac{h-1}{2}=\;\frac{k+2}{-3}=-\;\frac{26}{13}

\;\Rightarrow h=-3, k=4

Centre of circle

C(-3,4)

C:(x+3)^2+(y-4)^2=3^2

Co-ordinate of

A(0,4)

As

O A

is parallel to x-axis length of are

\;AB=\;\frac{1}{6} \times \;\text{ perimeter }\;

\;=\;\frac{1}{6} \times 2 \pi \times 3

\;\theta=\;\frac{\pi}{180} \times 3 = \pi

\;\Rightarrow\;\; \theta = 60^{\circ}

\;\;\text{ In }\; \triangle OPB

\;\sin\;30^{\circ}=\;\frac{PB}{OB}

\;\;\frac{PB}{3}=\;\frac{1}{2}

\;\Rightarrow\;\; PB=\;\frac{-3}{2}

\;\cos 30^{\circ}=\;\frac{OP}{3}

\;\Rightarrow OP=\;\frac{3\sqrt{3}}{2}

\;B \left(\;\frac{-3}{2}, 4-\;\frac{3\sqrt{3}}{2}\right)

\;\beta-\sqrt{3}\alpha=4-\;\frac{3\sqrt{3}}{2}+\;\frac{3\sqrt{3}}{2}=4

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