The angle between circles x²+y²+2 x+4 y+1=0 and x²+y²-2 x+6 y-3=0 is:

Correct Answer is : Cos−1(313){Cos}^{-1}≤ft({3}{{13}})Cos−1(133)

Correct Answer is : Cos−1(313){Cos}^{-1}≤ft({3}{{13}})Cos−1(133)

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AP EAPCET 25-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 55 of 160

Marks: +1, -0

x^{2}+y^{2}+2 x+4 y+1=0

x^{2}+y^{2}-2 x+6 y-3=0

is:

Solution: 👈: Video Solution

Circles are

c_{1}, x^{2}+y^{2}+2 x+4 y+1=0

c_{2}, x^{2}+y^{2}-2 x+6 y-3=0

On comparing with general form of circles

x^{2}+y^{2}+2 g x+2 f y+c=0

where, centre

=(-g,-f)

and radius

=\sqrt{g^{2}+f^{2}-c}

∴

c_{1}=(-1,-2)

and

r_{1}=\sqrt{1+4-1}=2

c_{2}=(1,-3)

and

r_{2}=\sqrt{1+9+3}=\sqrt{13}

∴

d=c_{1} c_{2}=\sqrt{(-1-1)^{2}+(-2+3)^{2}}=\sqrt{4+1}

=\sqrt{5}

\therefore \cos \theta = \left|\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\right| = \left|\frac{5-4-13}{2 \times 2 \times \sqrt{13}}\right| = \left|\frac{-3}{\sqrt{13}}\right|

⇒

\cos \theta = \frac{3}{\sqrt{13}}

∴

\theta = \cos^{-1}\left(\frac{3}{\sqrt{13}}\right)

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