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接点を(x1,y1)とすると、
(x1-1)²+(y1+3)²=10…①
(x1-1)(x-1)+(y1+3)(y+3)=10が(3、1)を通るから
2(x1-1)+4(y1+3)=10
x1+2y1=0…②
①②より、
(-2y1-1)²+(y1+3)²=10
5y1²+10y1=0
y1(y1+2)=0
y1=0、-2
y1=0よりx1=0
y1=-2よりx1=4

でどうでしょうか?

とまと

ありがとうございます!!

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