Solveeit Logo

Question

Mathematics Question on circle

If the circles x2+y2+6x+8y+16=0x^2+y^2+6 x+8 y+16=0 and x2+y2+2(33)x+x+2(46)yx^2+y^2+2(3-\sqrt{3}) x+x+2(4-\sqrt{6}) y = k +6 \sqrt{3}+8 \sqrt{6}, k &gt0; touch internally at the point P(α,β)P(\alpha, \beta), then (α+3)2+(β+6)2(\alpha+\sqrt{3})^2+(\beta+\sqrt{6})^2 is equal to _______

Answer

The correct answer is 25
C1​(−3,−4)
r1​=25−16​=3
C2​=(−3+3​,−4+6​)
r2​=34+k​
C1​C2​=∣r1​−r2​∣
C1​C2​=3+6​=3
3=∣3−34+k​∣⇒k=2
r2​=6

(α,β)=(−3​−3,−4−6​)
(α+3​)2+(β+6​)2=9+16=25