JEE MAIN - Mathematics (2003 - No. 49)
The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is :
$${x^2}\, + \,{y^2} - \,2x\, + \,2y\,\, = \,62$$
$${x^2}\, + \,{y^2} + \,2x\, - \,2y\,\, = \,62$$
$${x^2}\, + \,{y^2} + \,2x\, - \,2y\,\, = \,47$$
$${x^2}\, + \,{y^2} - \,2x\, + \,2y\,\, = \,47$$
Explanation
$$\pi {r^2} = 154 \Rightarrow r = 7$$
For center on solving equation
$$2x - 3y = 5\& 3x - 4y = 7$$
we get $$x = 1,\,y = - 1$$
$$\therefore$$ center $$=(1,-1)$$
Equation of circle,
$${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2}$$
$${x^2} + {y^2} - 2x + 2y = 47$$
For center on solving equation
$$2x - 3y = 5\& 3x - 4y = 7$$
we get $$x = 1,\,y = - 1$$
$$\therefore$$ center $$=(1,-1)$$
Equation of circle,
$${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2}$$
$${x^2} + {y^2} - 2x + 2y = 47$$
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