The value of n, for which the circle \( \Large x^{2}+y^{2}+2nx+6y+1=0 \) intersects the circle \( \Large x^{2}+y^{2}+4x+2y=0 \) orthogonally is:

A) \( \Large \frac{11}{8} \)

B) -1

C) \( \Large \frac{-5}{4} \)

D) \( \Large \frac{5}{2} \)

Correct answer:
C) \( \Large \frac{-5}{4} \)

Description for Correct answer:

Equations of circles are

\( \Large x^{2}+y^{2}+2nx+6y+1=0 \)

and \( \Large x^{2}+y^{2}+4x+2y=0 \)

Since, the circles cut orthogonally

\( \Large \therefore\ 2gg+2ff = c+c \)

=> \( \Large 2n \times 2+6 \times 1 = 1+0 \)

=> \( \Large 4h+6 = 1 \)

=> \( \Large n = \frac{-5}{4} \)



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