Geo4.12

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Find the equation to the circle passing through (0,0) and concentric with x^{2}+y^{2}-2x-3y-4=0\,

Concentric means that the both the circles will have the same centre.

Hence the centre of the circle is

\left({\frac  {-(-2)}{2}},{\frac  {-(-3)}{2}}\right)\,

\left(1,{\frac  {3}{2}}\right)\,

The radius of the circle with the centre above and passing through (0,0) is

{\sqrt  {1+{\frac  {9}{4}}}}\,

{\sqrt  {{\frac  {13}{4}}}}\,

Now the equation of the circle is

(x-1)^{2}+(y-{\frac  {3}{2}})^{2}={\frac  {13}{4}}\,

x^{2}+1-2x+y^{2}+{\frac  {9}{4}}-3y={\frac  {13}{4}}\,

4x^{2}+4y^{2}-2x-3y+10=13\,

Hence the equation of the circle is

4x^{2}+4y^{2}-2x-3y-3=0\,


Main Page:Geometry:Circles