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Question

Find the equation of the circle passing through the points of intersection of the line $$x+3y=0$$ and $$2x-7y=0$$ and whose centre  is the point of intersection of the lines $$x+y+1=0$$ and $$x-2y+4=0$$.


Solution

$$\begin{array}{l} The\, po{ { int } }\, of\ intersection\, of\, the\, line\, x+3y=0 \\ and\, 2x-7y=0\, \, is\, \left( { 0,\, 0 } \right)  \\ Let\, \left( { h,\, k\,  } \right) \, be\, the\, centre\, of\, circle\, with\, radius\, a. \\ thus,\, its\, equation\, will\, be\, { \left( { x-h } \right) ^{ 2 } }+{ \left( { y-k } \right) ^{ 2 } }={ a^{ 2 } } \\ The\, po{ { int } }\, of\, the\, { { int } }ersec  tion\, of\, the\, line\, x+y+1=0\, \, and\, x-2y+4=0\, is\, \left( { -2,\, 1 } \right)  \\ \therefore h=-2,\, k=1 \\ \therefore Equation\, of\, the\, required\, circle={ \left( { x+2 } \right) ^{ 2 } }+{ \left( { y-1 } \right) ^{ 2 } }={ a^{ 2 } }\, \, \, \, \, \, ----\left( 1 \right)  \\ also,\, equation\, \left( 1 \right) \, passes\, through\, \left( { 0,\, 0 } \right)  \\ \therefore { \left( { 0+2 } \right) ^{ 2 } }+{ \left( { 0-1 } \right) ^{ 2 } }={ a^{ 2 } } \\ 4+1={ a^{ 2 } }\therefore a=\sqrt { 5 } \, \, \, \, \left( { \therefore a>0 } \right)  \\ substituting\, the\, values\, fo\, a\, in\, equation\, \left( 1 \right)  \\ { \left( { x+2 } \right) ^{ 2 } }+{ \left( { y-1 } \right) ^{ 2 } }=5 \\ { x^{ 2 } }+4+4x+{ y^{ 2 } }+1-2y=5 \\ { x^{ 2 } }+4x+{ y^{ 2 } }-2y=0 \\ Hence,\, It\, is\, the\, required\, answer. \end{array}$$

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