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Circle

Section: Mathematics

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Question : 23 of 89

Marks: +1, -0

The radius of the circle passing through the points of intersection of the circles

x^2+y^2+2 x+4 y+1=0, x^2+y^2-2 x-4 y-4=0

and intersecting the circle

x^2+y^2=6

orthogonally is

[AP EAPCET 23 May 2025 S1]

$\sqrt{19}$
5
$\sqrt{39}$
$4$

Solution:

The equation of a circle passing through the intersection of two circles

\;x^2+y^2+2 x+4 y+1=0 \;\text{ and }\;

\;x^2+y^2-2 x-4 y-4=0 \;\text{ is }\;

\; (x^2+y^2+2 x+4 y+1)

\;+\lambda (x^2+y^2-2 x-4 y-4)=0, \;\text{ where }\; \lambda

is a parameter.

\Rightarrow (1+\lambda) x^2\;+(1+\lambda) y^2+(2-2\lambda)+(4-4\lambda) y+1-4\lambda=0

\;\Rightarrow x^2+y^2+\;\frac{2(1-\lambda)}{(1+\lambda)} x+\;\frac{4(1-\lambda)}{(1+\lambda)} y+\;\frac{1-4\lambda}{1+\lambda}=0 \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)

Applying the orthogonal condition,

2 \left(\;\frac{1-\lambda}{1+\lambda}\right) \times 0\;+2 \times \;\frac{2(1-\lambda)}{(1+\lambda)} \times 0

\;=\;\frac{1-4\lambda}{1+\lambda}-6

\Rightarrow \;\; \;\frac{1-4\lambda}{1+\lambda}\;=6

\Rightarrow \;\; 6+6\lambda\;=1-4\lambda \Rightarrow \lambda=\;\frac{-1}{2}

Substituting

\lambda=\;\frac{-1}{2}

in Eq. (i) and simplifying we get

x^2+y^2+6 x+12 y+6=0

∴ Radius of this circle

=\sqrt{3^2+6^2-6}=\sqrt{39}

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