If the circles \(x^2+y^2+6 x+8 y+16=0\) and \(x^2+y^2+2(3-\sqrt{3}) x+2(4-\sqrt{6}) y=k+6 \sqrt{3}+8 \sqrt{6}\), \(k>0\), touch internally at the point \(P(\alpha, \beta)\), then \((\alpha+\sqrt{3})^2+(\beta+\sqrt{6})^2\) is equal to 📲PW App Link - bit.ly/YTAI_PWAP 🌐PW Website - www.pw.live