JEE MAIN - Mathematics (2006 - No. 27)
If the lines $$3x - 4y - 7 = 0$$ and $$2x - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi $$ square units, the equation of the circle is :
$$\,{x^2} + {y^2} + 2x\, - 2y - 47 = 0\,$$
$$\,{x^2} + {y^2} + 2x\, - 2y - 62 = 0\,$$
$${x^2} + {y^2} - 2x\, + 2y - 62 = 0$$
$${x^2} + {y^2} - 2x\, + 2y - 47 = 0$$
Explanation
Point of intersection of $$3x - 4y - 7 = 0$$ and
$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the
circle and radius $$=7$$
$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$
$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$
$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the
circle and radius $$=7$$
$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$
$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$
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