JEE MAIN - Mathematics (2002 - No. 53)
If the chord y = mx + 1 of the circle $${x^2}\, + \,{y^2} = 1$$ subtends an angle of measure $${45^ \circ }$$ at the major segment of the circle then value of m is :
$$2\, \pm \,\sqrt 2 \,\,$$
$$ - \,2\, \pm \,\sqrt 2 \,$$
$$- 1\, \pm \,\sqrt 2 \,\,$$
none of these
Explanation
Equation of circle $${x^2} + {y^2} = 1 = {\left( 1 \right)^2}$$
$$ \Rightarrow {x^2} + {y^2} = {\left( {y - mx} \right)^2}$$
$$ \Rightarrow {x^2} = {m^2}{x^2} - 2\,\,mxy;$$
$$ \Rightarrow {x^2}\left( {1 - {m^2}} \right) + 2mxy = 0.$$
Which represents the pair of lines between which the angle is $${45^ \circ }.$$
$$\tan 45 = \pm {{2\sqrt {{m^2} - 0} } \over {1 - {m^2}}} = {{ \pm 2m} \over {1 - {m^2}}};$$
$$ \Rightarrow 1 - {m^2} = \pm 2m$$
$$ \Rightarrow {m^2} \pm 2m - 1 = 0$$
$$ \Rightarrow m = {{ - 2 \pm \sqrt {4 + 4} } \over 2}$$
$$ = {{ - 2 \pm 2\sqrt 2 } \over 2}$$
$$ = - 1 \pm \sqrt 2 .$$
$$ \Rightarrow {x^2} + {y^2} = {\left( {y - mx} \right)^2}$$
$$ \Rightarrow {x^2} = {m^2}{x^2} - 2\,\,mxy;$$
$$ \Rightarrow {x^2}\left( {1 - {m^2}} \right) + 2mxy = 0.$$
Which represents the pair of lines between which the angle is $${45^ \circ }.$$
$$\tan 45 = \pm {{2\sqrt {{m^2} - 0} } \over {1 - {m^2}}} = {{ \pm 2m} \over {1 - {m^2}}};$$
$$ \Rightarrow 1 - {m^2} = \pm 2m$$
$$ \Rightarrow {m^2} \pm 2m - 1 = 0$$
$$ \Rightarrow m = {{ - 2 \pm \sqrt {4 + 4} } \over 2}$$
$$ = {{ - 2 \pm 2\sqrt 2 } \over 2}$$
$$ = - 1 \pm \sqrt 2 .$$
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