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Circle

Section: Mathematics

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Question : 39 of 89

Marks: +1, -0

If the acute angle between the circles

S \equiv x^2+y^2+2 k x+4 y-3=0

S^1 \equiv x^2+y^2-4 x+2 k y+9=0

\cos^{-1}\left(\frac{3}{8}\right)

and the centre of

S^1=0

lies in the first quadrant, then the radical axis of

S=0

S^1=0

[AP EAPCET 21 May 2025 S1]

$x-5y+6=0$
$x-5y-4=0$
$5x-y-6=0$
$5x-y-4=0$

Solution:

S = x^2+y^2+2 k x+4 y-3=0

C_1=(-k,-2) \text{ radius } r_1=\sqrt{k^2+4+3}

S' = x^2+y^2-4 x+2 k y+9=0

C_2=(2,-k) \text{ radius } =\sqrt{4+k^2-9}

=\sqrt{k^2-5}

Now, angle between two circles

S=0

and

S'=0

is

\cos(180^{\circ}-\theta)=\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}

where

r_1, r_2

are the radii and

d

is the distance between their centres.

=\frac{k^2+7+k^2-5-(k+2)^2-(k-2)^2}{2 \sqrt{k^2+7} \sqrt{k^2-5}}

\Rightarrow -\cos\theta = \frac{2k^2+2-(2k^2+8)}{2 \sqrt{k^2+7} \sqrt{k^2-5}}

= \frac{-6}{2 \sqrt{k^2+7} \sqrt{k^2-5}}

\Rightarrow \cos\theta = \frac{3}{\sqrt{k^2+7} \sqrt{k^2-5}} = \frac{3}{8}

\Rightarrow (k^2+7)(k^2-9)=64

\Rightarrow k^4+2k^2-99=0

\Rightarrow (k^2+11)(k^2-9)=0 \Rightarrow k^2=9

\Rightarrow k=\pm 3

Centre of

S'=0

lies in I quadrant

[\therefore k=-3]

Now, radical axis of

S=0

and

S'=0

is

\Rightarrow x(2k+4)+y(4-2k)-12=0 \\ -2x+10y-12=0 \\ x-5y+6=0

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