Question Number 68710 by peter frank last updated on 15/Sep/19
Commented by Prithwish sen last updated on 15/Sep/19
$$\mathrm{x}+\mathrm{y}=\mathrm{0}…….\left(\mathrm{i}\right) \\ $$$$\mathrm{x}−\mathrm{y}=−\mathrm{4}……\left(\mathrm{ii}\right) \\ $$$$\mathrm{solving}\:\mathrm{x}=−\mathrm{2},\mathrm{y}=\mathrm{2} \\ $$$$\therefore\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{center}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{is}}\:\left(−\mathrm{2},\mathrm{2}\right) \\ $$$$\therefore\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{radius}}\:=\mid\:\frac{−\mathrm{2}−\mathrm{2}}{\:\sqrt{\mathrm{2}}}\mid=\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\because\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{perpendicular}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{tangent}} \\ $$$$\therefore\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{cicle}} \\ $$$$\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{y}}−\mathrm{2}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{x}}−\mathrm{4}\boldsymbol{\mathrm{y}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
Commented by peter frank last updated on 15/Sep/19
$${correct}\:{sir} \\ $$
Commented by peter frank last updated on 15/Sep/19
$${how}\:{about}\:{Qn}\:\mathrm{2} \\ $$