Suppose that the circle \(x^2+y^2+2 g x+2 f y+c=0\) has its centre on \(2 x+3 y-7=0\) and cuts the circles…
Suppose that the circle \(x^2+y^2+2 g x+2 f y+c=0\) has its centre on \(2 x+3 y-7=0\) and cuts the circles \(x^2+y^2-4 x-6 y+11=0\) and \(x^2+y^2-10 x-4 y+21=0\) orthogonally. Then \(5 g-10 f+3 c=\)
0
1
3
9
Solution
The equation of given circle
\(x^2+y^2+2 g x+2 f y+c=0\) ...(i)
having centre \((-g,-f)\) lying on the line
\(\begin{aligned}
& 2 x+3 y-7=0 \\
& \text{so } 2 g+3 f+7=0 \quad \ldots (ii)
\end{aligned}\)
Since circle (i) cuts the given circles
\(\quad \begin{aligned}
x^2+y^2-4 x-6 y+11 & =0 \\
\text {and } x^2+y^2-10 x-4 y+21 & =0
\end{aligned}\)
orthogonally
\(\begin{array}{lc}
\text { so, } & 2 g(-2)+2 f(-3)=c+11 \\
\Rightarrow & 4 g+6 f+c+11=0 \quad \ldots (iii) \\
\text { and } & 2 g(-5)+2 f(-4=c+21 \\
\Rightarrow & 10 g+4 f+c+21=0 \quad \ldots (iv)
\end{array}\)
From Eqs. (iii) and (iv), we get
\(6 g-2 f+10=0 \quad \ldots (v) \)
From Eq. (ii) and (v), we get
\(\begin{gathered}
11 f+11=0 \Rightarrow f=-1 \\
\text{So, } g=-2 \Rightarrow c=3
\end{gathered}\)
\(\begin{aligned}
& \therefore 5 g-10 f+3 c=5(-2)-10(-1)+3(3) \\
& =-10+10+9=9
\end{aligned}\)
Hence, option (4) is correct.