Question

For $\alpha ,\beta \in R,$ if the circles $x^2+y^2+(3+\sin \beta )x+\left(2\cos \alpha \right)y=0$ and $x^2+y^2+\left(2\cos \alpha \right)x+2cy=0$ touch each other, then the maximum value of $c$ is

• A. $6$
• B. $1$
• C. $5$
• D. $9$

Answer 1

The correct option is B $1$

Let $S_1:x^2+y^2+(3+\sin \beta )x+\left(2\cos \alpha \right)y=0$
$S_2:x^2+y^2+\left(2\cos \alpha \right)x+2cy=0$
Both the circles are passing through the origin $(0,0).$
Equation of tangent at $(0,0)$ to $S_1$ is $T=0$
i.e., $(3+\sin \beta )x+\left(2\cos \alpha \right)y=0\dots \left(1\right)$
Equation of tangent at $(0,0)$ to $S_2$ is $T=0$
i.e., $\left(2\cos \alpha \right)x+2cy=0\dots \left(2\right)$

Since $S_1$ and $S_2$ touch each other, hence equation $\left(1\right)$ and $\left(2\right)$ must be identical.
Comparing equation $\left(1\right)$ and $\left(2\right)$, we get
$\frac{3+\sin \beta }{2\cos \alpha }=\frac{2\cos \alpha }{2c}\Rightarrow c=\frac{2{\cos }^2\alpha }{3+\sin \beta }$
So, we get maximum value of $c$ when $\sin \beta =-1$ and $\cos \alpha =1$
$\therefore c_{max}=1$