JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

1 Medium

1 of 1

Application +4 | -1

The limiting points of coaxial-system determined by the circles

{x^2} + {y^2} + 5x + y + 4 = 0

and

{x^2} + {y^2} + 10x - 4y - 1 = 0

are

View answer and explanation

Correct option: B

\left( {0, - 3} \right)

and

\left( { - 2, - 1} \right)

Option B (

\left( {0, - 3} \right)

and

\left( { - 2, - 1} \right)

) is correct because these are the limiting points found by setting the radius of the coaxial system of circles to zero.

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