If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also…
If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also the line \(\frac{x}{5}+\frac{y}{12}=1\) is \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0\), then \(\lambda=\)
3
10
15
-2
Solution
Given, equation of circle is
\(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0\)...(i)
Here, centre is \((-\lambda, \lambda)\) and radius \(=\lambda\)
Since, circle (i) touches \(12 x+5 y-60=0\)
\(\begin{array}{ll}
\therefore & r=d \\
\Rightarrow & \lambda=\frac{|-7 \lambda-60|}{\sqrt{144+25}} \\
\Rightarrow & 13 \lambda=|-7 \lambda-60| \\
\Rightarrow & 7 \lambda+60=13 \lambda \text { or }-7 \lambda-60=13 \lambda \\
\Rightarrow & 6 \lambda=60 \text { or } 20 \lambda=-60 \\
\Rightarrow & \lambda=10 \text { or } \lambda=-3 \\
\Rightarrow & \lambda=10 \quad (\because \lambda \neq-3)
\end{array}\)