# Find the equation of the circle which touches the axes and whose centre lies on x-2y = 3.

$Let\;us\;assume\;the\;circle\;touches\;the\;axes\;at\;(a,0)\;and\;(0,a)\;and\\* we\;get\;the\;radius\;to\;be\;|a|\\*We\;get\;the\;centre\;of\;the\;circle\;as\; (a,a).\\*This\;point\;lies\;on\;the\;line\;x-2y=3\\*\Rightarrow a-2(a)=3\\* \Rightarrow a=3\\*\Rightarrow a=-3\\*Centre=(a,a)=(-3,-3)\;and\;radius\; of\;the\;circle(r)=|-3|=3\\*We\;have\;circle\;with\;centre\;(-3,-3)\; and\;having\;radius\;3.\\*We\;know\;that\;the\;equation\;of\;the\;circle\; with\;centre\;(p,q)\;and\;having\\*radius,r\;is\;given\;by:(x-p)^2+(y-q)^2=r^2 \\*Now\;by\;substituting\;the\;values\;in\;the\;equation,\;we\;get\\*\Rightarrow (x-(-3))^2+(y-(-3))^2=32\\*\Rightarrow (x+3)^2+(y+3)^2=9\\*\Rightarrow x^2+6x+9+y^2+6y+9=9\\*\Rightarrow x^2+y^2+6x+6y+9=0\\* \therefore The\;equation\;of\;the\;circle\;is\;x^2+y^2+6x+6y+9=0.$

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