JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

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Analyzing +4 | -1

{x^2} + {y^2} - 6x - 6y + 4 = 0,{x^2} + {y^2} - 2x - 4y + 3 = 0,

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

. If the radical centre of the above three circles exists, then which of the following cannot be the value of

k

View answer and explanation

Correct option: C

Option C (5) is correct because the radical center exists only when the determinant of the coefficients of the radical axes is non-zero, leading to

k \ne 5

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x = 1

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

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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

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

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5 x+b y-a=0

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x + 2y - 3 = 0

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

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