Answer :
[tex]|PA|=\sqrt{(1-x)^2+(-2-y)^2}\\\\|PB|=\sqrt{(4-x)^2+(-3-y)^2}\\\\|PA|=2|PB|\\\\\sqrt{(1-x)^2+(-2-y)^2}=2\sqrt{(4-x)^2+(-3-y)^2}\\\\(1-x)^2+(-2-y)^2=4[(4-x)^2+(-3-y)^2]\\\\1-2x+x^2+4+4y+y^2=4(16-8x+x^2+9+6y+y^2)[/tex]
[tex]x^2+y^2-2x+4y+5=4(x^2+y^2-8x+6y+25)\\\\x^2+y^2-2x+4y+5=4x^2+4y^2-32x+24y+100\\\\x^2-4x^2-2x+32x+y^2-4y^2+4y-24y+5-100=0\\\\-3x^2+30x-3y^2-20y-95=0\ \ \ \ /:(-3)\\\\x^2-10x+y^2+\frac{20}{3}y+\frac{95}{3}=0\\\\x^2-2x\cdot5+5^2-5^2+y^2+2y\cdot\frac{10}{3}+\left(\frac{10}{3}\right)^2-\left(\frac{10}{3}\right)^2=-\frac{95}{3}[/tex]
[tex](x-5)^2+(y+\frac{10}{3})^2=-\frac{95}{3}+\frac{100}{9}+25\\\\(x-5)^2+(y+\frac{10}{3})^2=-\frac{285}{9}+\frac{100}{9}+\frac{225}{9}\\\\(x-5)^2+(y+\frac{10}{3})^2=\frac{40}{9}\\\\it's\ the\ circle\\\\center:(5;-\frac{10}{3})\\\\radius:r=\sqrt\frac{40}{9}=\frac{\sqrt{4\cdot10}}{\sqrt9}=\frac{2\sqrt{10}}{3}[/tex]
[tex]x^2+y^2-2x+4y+5=4(x^2+y^2-8x+6y+25)\\\\x^2+y^2-2x+4y+5=4x^2+4y^2-32x+24y+100\\\\x^2-4x^2-2x+32x+y^2-4y^2+4y-24y+5-100=0\\\\-3x^2+30x-3y^2-20y-95=0\ \ \ \ /:(-3)\\\\x^2-10x+y^2+\frac{20}{3}y+\frac{95}{3}=0\\\\x^2-2x\cdot5+5^2-5^2+y^2+2y\cdot\frac{10}{3}+\left(\frac{10}{3}\right)^2-\left(\frac{10}{3}\right)^2=-\frac{95}{3}[/tex]
[tex](x-5)^2+(y+\frac{10}{3})^2=-\frac{95}{3}+\frac{100}{9}+25\\\\(x-5)^2+(y+\frac{10}{3})^2=-\frac{285}{9}+\frac{100}{9}+\frac{225}{9}\\\\(x-5)^2+(y+\frac{10}{3})^2=\frac{40}{9}\\\\it's\ the\ circle\\\\center:(5;-\frac{10}{3})\\\\radius:r=\sqrt\frac{40}{9}=\frac{\sqrt{4\cdot10}}{\sqrt9}=\frac{2\sqrt{10}}{3}[/tex]
