Find the value of \(m+n\), if the circumference of the circle \(x^2+y^2+8 x+8 y-m=0\) is bisected by the…
Find the value of \(m+n\), if the circumference of the circle \(x^2+y^2+8 x+8 y-m=0\) is bisected by the circle \(x^2+y^2-2 x+4 y+n=0\).
-56
56
50
-34
Solution
Let \(S_1 \equiv x^2+y^2+8 x+8 y-m=0\)
\(S_2=x^2+y^2-2 x+4 y+n=0\)
Equation of common chord
\(\begin{aligned}
\Rightarrow & & S_1-S_2 & =0 \\
\Rightarrow & & 10 x+4 y-(m+n) & =0 \\
\Rightarrow & & 10 x+4 y & =(m+n)
\end{aligned}\)
Centre of bisected circle is \((-4,-4)\) which will lie on common chord
So, \(\quad\{(m+n)=-56\}\)