[tex]A(4, 1)\\k: x-y=0 \to y=x\\S \in k \to S(x, x)[/tex]
Okregi sa styczne wtedy, kiedy odleglosc od srodka okregu do prostej jest rowna promieniowi okregu.
[tex]|SA|=d_{S,l}[/tex]
[tex]r=|SA|=\sqrt{(x-4)^2+(x-1)^2}[/tex]
[tex]d_{S, l}=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}[/tex]
[tex]d_{S, l}=\frac{|0x+1x-5|}{\sqrt{0^2+1^2}}=\frac{|x-5|}{1}=|x-5|[/tex]
[tex]|x-5|=\sqrt{(x-4)^2+(x-1)^2}\\x-5=\sqrt{(x-4)^2+(x-1)^2} /^2\\(x-5)^2=(x-4)^2+(x-1)^2\\x=2\sqrt2, x=-2\sqrt2\\\\-(x-5)=\sqrt{(x-4)^2+(x-1)^2}/ ^2\\(5-x)^2=(x-4)^2+(x-1)^2\\x=2\sqrt2, x=-2\sqrt2[/tex]
[tex]S=(2\sqrt2, 2\sqrt2) \text{ lub } S=(-2\sqrt2, -2\sqrt2)[/tex]
1)
Dla [tex]S=(2\sqrt2, 2\sqrt2)\\[/tex]
[tex]r=\sqrt{(2\sqrt2-4)^2+(2\sqrt2-1)^2}=\sqrt{33-20\sqrt2}[/tex]
Rownanie okregu:
[tex](x-2\sqrt2)^2+(y-2\sqrt2)^2=33-20\sqrt2[/tex]
2)
Dla [tex]S=(-2\sqrt2, -2\sqrt2)[/tex]
[tex]r=\sqrt{(-2\sqrt2-4)^2+(-2\sqrt2-1)^2}=\sqrt{33+20\sqrt2}[/tex]
Rownanie okregu:
[tex](x+2\sqrt2)^2+(y+2\sqrt2)^2=33+20\sqrt2[/tex]
Odp. Rownanie tego okregu to: [tex]\underline{(x-2\sqrt2)^2+(y-2\sqrt2)^2=33-20\sqrt2}[/tex] lub [tex]\underline{(x+2\sqrt2)^2+(y+2\sqrt2)^2=33+20\sqrt2}[/tex]