Given a circle passing through point P (2, - 1), the center of the circle is on L1: y + 2x = 0 and tangent to the line L2: x-y-1 = 0, the equation of the circle is obtained | Math enthusiast | Mathematical art

Given a circle passing through point P (2, - 1), the center of the circle is on L1: y + 2x = 0 and tangent to the line L2: x-y-1 = 0, the equation of the circle is obtained

Let the coordinates of the center o be (a, b),
b=-2a
((a-2)^2)+((-2b+1)^2)=([(a+2a-1)/√((1^2)+((-1)^2))]^2)
∴a1=1 a2=9
O1P=√(2)
O2P=√(338)
The equation of circle O is:
((x-1)^2)+((y+2)^2)=2
((x-9)^2)+((y+18)^2)=338

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