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If the circle \( \Large x^{2}+y^{2}+6x-2y+k=0 \) bisects the circumference of the circle \( \Large x^{2}+y^{2}+2x-6y-15=0 \) then k is equal to:

A) 21
B) -21
C) 23
D) -23

Correct Answer:




D) -23

Description for Correct answer:
The condition for a circle bisecting the circumference of the second circle is:

\( \Large 2g_{2} \left(g_{1}-g_{2}\right)+2f_{2} \left(f_{1}-f_{2}\right)=c_{1}-c_{2} \)

=> \( \Large 2 \left(1\right) \left(3-1\right)+2 \left(-3\right) \left(-1+3\right)=k+15 \)

=> \( \Large -8 = k + 15 => k = -23 \)

Part of solved Circles questions and answers : >> Elementary Mathematics >> Circles

Comments

Similar Questions

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