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If circle \mathrm{x^2+y^2+4 x+8 y-29=0} bisects circumference of the circle \mathrm{x^2+y^2+2 x+3 y+k=0}, then \mathrm{k} is equal to
 

Option: 1

50

 


Option: 2

\frac{55}{2}
 


Option: 3

-\frac{67}{2}
 


Option: 4

-\frac{77}{2}


Answers (1)

best_answer

Centre of 2^{\text {nd }} circle i.e., \mathrm{\left(-1,-\frac{3}{2}\right)}, must satisfy equation of common chord.

Equation of common chord is \mathrm{2 x+5 y-29-k=0}

\mathrm{\Rightarrow \quad -2-\frac{15}{2}-29-k=0 }

\mathrm{\Rightarrow k=-\frac{77}{2} }

Hence option 4 is correct.



 

Posted by

avinash.dongre

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