Question

If the circles $x ^2 + y ^2 =9$ and $x^2 + y^ 2 + 2\alpha x + 2y +1 = 0$ touch each other internally, then $\alpha$ =

• A. $\pm \frac {4}{3}$
• B. $1$
• C. $ \frac {4}{3}$
• D. \(- \frac {4}{3}\)

Answer 1

The Correct Option is A

The correct option is(A): \(\pm \frac {4}{3}\).

Centers and radii of the given circles \(x^{2}+y^{2}=9\) 
and \(x^{2}+y^{2}+2 a x+2 y+1=0\) is \(C_{1}(0,0), r_{1}=3\)
and \(C_{2}(-\alpha, 1)\) and \(r_{2}=\sqrt{\alpha^{2}+1-1}=|\alpha|\)
Since, two circles touch internally, 
\(\therefore C_{1} C_{2}=r_{1}-r_{2}\)
\(\Rightarrow \,\,\,,\sqrt{\alpha^{2}+1^{2}}=3-|\alpha|\)
\(\Rightarrow \,\,\,\alpha^{2}+1=9+\alpha^{2}-6|\alpha|\)
\(\Rightarrow \,\,\,\,6|\alpha|=8\)
\(\Rightarrow|\alpha|=\frac{4}{3}\)
\(\Rightarrow \,\,\,\alpha=\pm \frac{4}{3}\)