JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

1 Easy

1 of 1

Application +4 | -1

The value of

\lambda

, for which the circle

{x^2} + {y^2} + 2\lambda x + 6y + 1 = 0

intersects the circle

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

orthogonally, is

View answer and explanation

Correct option: C

\frac{{ - 5}}{4}

Option C (

\frac{{ - 5}}{4}

) is correct because it satisfies the orthogonality condition

2g_1g_2 + 2f_1f_2 = c_1 + c_2

for the given circles.

More Questions from SYSTEM OF CIRCLES

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x = 1

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

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

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

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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