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  • If the circumference of the circle <img alt="\mathrm{x^2+y^2+8 x+8 y-b=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2&plus;8%20x&plus;8%20y-b%3D0%

If the circumference of the circle \mathrm{x^2+y^2+8 x+8 y-b=0} is bisected by the circle \mathrm{x^2+y^2-2 x+4 y+a=0}, then \mathrm{a+b=}

 

Option: 1

50


Option: 2

56


Option: 3

-56


Option: 4

-34


Answers (1)

best_answer

 Equation of radical axis (i.e. common chord) of the two circles is \mathrm{S_1-S_2=0}

\mathrm{\Rightarrow \quad 10 x+4 y-a-b=0}

Centre of first circle is \mathrm{C_1(-4,-4)}.
Since second circle bisects the circumference of the first circle, therefore, centre \mathrm{C_1(-4,-4)} of the first circle must lie on the common chord (i)

\mathrm{\therefore-40-16-a-b=0 \Rightarrow a+b=-56 }

Posted by

himanshu.meshram

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