{х²+у²=10
{х²+у=р
р-?
х²+у²-х²-у=(10-р)
у²-у+(р-10)=0
D=(-(-1))²-4×1×(p-10)=1+4(p-10)=1-4p+40=-4p+41.
D=0
-4p+41=0
-4p=-41|÷(-4)
p=41/4=10(1/4)=10,25
y²-y-((41/4)-10)=0
y²-y-1/4=0
D=(-(-1))²-4×1×(-(1/4))=1+1=2
y1=(-(-1)-√2)/2×1=(1-√2)/2=(1-1,41)/2=-0,205
y2=(-(-1)+√2)/2×1=(1+√2)/2=(1+1,41)/2=1,205
(x1)²+y1=p
(x1)²-(-0,205)=10,25
(x1)²+0,205=10,25
(x1)²=10,25-0,205
(x1)²=10,045
(x1)1=-3,17
(x1)2=3,17
(x2)²+y2=p
(x2)²+1,205=10,25
(x2)²=10,25-1,205
(x2)²=9,045
(x2)1=-3,01
(x2)2=3,01
Ответ: р=10,25, (-3,17;-0,205), (3,17;-0,205), (-3,01;1,205), (3,01;1,205)