JEE MAIN - Mathematics (2005 - No. 47)
If the circles $${x^2}\, + \,{y^2} + \,2ax\, + \,cy\, + a\,\, = 0$$ and $${x^2}\, + \,{y^2} - \,3ax\, + \,dy\, - 1\,\, = 0$$ intersect in two ditinct points P and Q then the line 5x + by - a = 0 passes through P and Q for :
exactly one value of a
no value of a
infinitely many values of a
exactly two values of a
Explanation
$${s_1} = {x^2} + {y^2} + 2ax + cy + a = 0$$
$${s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0$$
Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is
given by $${s_1} - {s_2} = 0$$
$$ \Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$
Given that $$5x + by - a = 0$$ passes through $$P$$ and $$Q$$
$$\therefore$$ The two equations should represent the same line
$$ \Rightarrow {a \over 1} = {{c - d} \over b} = {{a + 1} \over { - a}}$$
$$ \Rightarrow a + 1 = - {a^2}$$
$${a^2} + a + 1 = 0$$
No real value of $$a.$$
$${s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0$$
Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is
given by $${s_1} - {s_2} = 0$$
$$ \Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$
Given that $$5x + by - a = 0$$ passes through $$P$$ and $$Q$$
$$\therefore$$ The two equations should represent the same line
$$ \Rightarrow {a \over 1} = {{c - d} \over b} = {{a + 1} \over { - a}}$$
$$ \Rightarrow a + 1 = - {a^2}$$
$${a^2} + a + 1 = 0$$
No real value of $$a.$$
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