6. What is the minimum value that the function $F(x ; y)=6 y+8 x-9$ 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 $F(x ; y)=\mathrm{M}$, then $\mathrm{M}=6 y+8 x-9$ - this is a straight line.

The condition for the minimum of the function is equivalent to the condition for the minimum of $\mathrm{M}$, under which the given line will be tangent to the circle. That is, the distance from the center of the circle to the line is equal to its radius.

$$
\frac{|30+40-9-M|}{\sqrt{36+64}}=5 \Rightarrow |61-M|=50 \Rightarrow M=11 \text{ or } M=111.
$$

## Answer: 11.