If the circles x2+y2+6x+8y+16=0 and x2+y2+2(3−3)x+2(4−6)y=k+63+86, k>0, touch internally at the point P(α,β), then (α+3)2+(β+6)2 is equal to \_\_\_\_\_\_\_\_.
Answer: 25
Solutions · 1
Official solution
First circle: centre (−3,−4), radius 3. Second: centre (3−3,6−4), radius 34+k (using (3−3)2+(4−6)2+63+86=34).
Distance between centres =3+6=3. Internal tangency: 3=∣34+k−3∣⇒34+k=6⇒k=2, so r2=6.
Unit vector from centre 2 to centre 1: (−33,−36). Tangency point is at distance r2 from centre 2 along this direction:
\[P=(\sqrt{3}-3,\sqrt{6}-4)+6\left(-\frac{\sqrt{3}}3,-\frac{\sqrt{6}}3\right)=(-\sqrt{3}-3,\,-\sqrt{6}-4).\]
\[(\alpha+\sqrt{3})^2+(\beta+\sqrt{6})^2=(-3)^2+(-4)^2=9+16=25.\]