6. What is the minimum value that the function $F(x ; y)=x^{2}+8 y+y^{2}+14 x-6$ can take, given that $x^{2}+y^{2}+25=10(x+y)$.

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# Solution.

$x^{2}+y^{2}+25=10(x+y) \Leftrightarrow (x-5)^{2}+(y-5)^{2}=5^{2}$ - this is a circle with center $(5 ; 5)$ and radius 5. Let $F(x ; y)=$ M, then $(x+7)^{2}+(y+4)^{2}=(M+71)$ - this is a circle with center $(-7 ;-4)$ and radius $(M+71)^{0.5}$.

Since the center of the second circle lies outside the first, the condition for the minimum of the function is equivalent to the condition for the minimum of M, at which the circles will intersect, and this is the point of tangency of the two circles. That is, the sum of the radii of the two circles should equal the distance between their centers. $\left(10+(M+71)^{0.5}\right)^{2}=(5+7)^{2}+(5+4)^{2}=>(M+71)^{0.5}=10=>\quad M=29$.

Answer: 29.