The equation of the circle which cuts the circles \(\begin{aligned} & S_1 \equiv x^2+y^2-4=0 \\ & S_2 \equiv…
The equation of the circle which cuts the circles
\(\begin{aligned}
& S_1 \equiv x^2+y^2-4=0 \\
& S_2 \equiv x^2+y^2-6 x-8 y+10=0 \\
& S_3 \equiv x^2+y^2+2 x-4 y-2=0
\end{aligned}\)
at the extremities of diameters of these circles is
\(x^2+y^2-4 x-6 y-4=0\)
\(x^2+y^2+4 x-4=0\)
\(x^2+y^2=25\)
\(x^2+y^2+x+y+1=0\)
Solution
Let the equation of required circle is
\(S \equiv x^2+y^2+2 g x+2 f y+c=0\)
and equation of given circles
\(\begin{aligned}
& S_1 \equiv x^2+y^2-4=0 \\
& S_2 \equiv x^2+y^2-6 x-8 y+10=0 \\
& S_3=x^2+y^2+2 x-4 y-2=0
\end{aligned}\)
\(\because\) Circle \(S=0\) cuts the circles \(S_1=0, S_2=0\) and \(S_3=0\) at the extermities of the diameters, so common chord of \(S=0\) and \(S_1=0\) passes through the centre of the circle \(S_1=0\), so \(c=-4\)
Similarly \((2 g+6) x+(2 f+8) y-14=0\) passes through \((3,4)\), so \(6 g+8 f+36=0\)
\(\Rightarrow \quad 3 g+4 f+18=0\)...(i)
and \((2 g-2) x+(2 f+4) y-2=0\), passes through \((-1,2)\), so
\(-2 g+4 f+8=0\)...(ii)
From Eqs. (i) and (ii), we get
\((g, f)=(-2,-3)\)
So equation of required circle is
\(x^2+y^2-4 x-6 y-4=0\)