JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

1 Easy

1 of 1

Application +4 | -1

The length of the common chord of the circles

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

and

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

View answer and explanation

Correct option: A

\sqrt {\frac{{15}}{2}}

Option A (

\sqrt {\frac{{15}}{2}}

) is correct because it represents the length of the common chord of the two circles.

More Questions from SYSTEM OF CIRCLES

Practice All Chapter MCQs →

x = 1

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

{x^2} + {y^2} + 8y - 4 = 0

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

x^2+y^2+3 x+2 y+1=0

x^2+y^2+3 x+4 y+2=0

{x^2} + {y^2} + 2ky + 2x + 6 = 0

{x^2} + {y^2} + 2ky + k = 0

k

x^2+y^2+2 a x+c y+a=0

x^2+y^2-3 a x+d y-1=0

P

Q

5 x+b y-a=0

P

Q

x + 2y - 3 = 0

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

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