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Circle

Section: Mathematics

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Question : 26 of 89

Marks: +1, -0

If the circle

S=x^2+y^2+2 g x+4 y+1=0

bisects the circumference of the circle

x^2+y^2-2 x-3=0

, then the radius of circle

S=0

[AP EAPCET 21 May 2025 S2]

5
$\sqrt{12}$
$25$
$12$

Solution:

We have the circle

S

S=x^2+y^2+2 g x+4 y+1=0

and the second circle be

S_1

,

S_1=x^2+y^2-2 x-3=0

\because

Centre

(1,0)

and radius

=2

Now, common chord of circles

S

and

S_1

is

\;S-S_1=0

\;(2 g+2) x+4 y+4=0

If a circle bisects the circumference of another circle, then common chord is a diameter of the second circle.

\;2 g+2+4=0 \Rightarrow g=-3

\because \;\; S=x^2+y^2-6 x+4 y+1=0

Centre

(3,-2)

Radius

=\sqrt{9+4-1}=\sqrt{12}

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