If the portion of the line lx + my = 1 fallin | vedclass

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(C) To find the angle subtended by the chord at the origin,we homogenize the equation of the circle ${x^2} + {y^2} = {a^2}$ using the line $lx + my =

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$4[{a^2}({l^2} + {m^2}) - 1] = {[{a^2}({l^2} + {m^2}) - 2]^2}$

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(C) To find the angle subtended by the chord at the origin,we homogenize the equation of the circle ${x^2} + {y^2} = {a^2}$ using the line $lx + my = 1$.${x^2} + {y^2} = {a^2}{(lx + my)^2}$${x^2} + {y^2} = {a^2}({l^2}{x^2} + {m^2}{y^2} + 2lmxy)$$({a^2}{l^2} - 1){x^2} + 2{a^2}lmxy + ({a^2}{m^2} - 1){y^2} = 0$This is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = {a^2}{l^2} - 1$,$H = {a^2}lm$,and $B = {a^2}{m^2} - 1$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{H^2 - AB}}{A + B} ight|$.Given $	heta = 45^\circ$,so $	an 45^\circ = 1$.$1 = \left| \frac{2\sqrt{({a^2}lm)^2 - ({a^2}{l^2} - 1)({a^2}{m^2} - 1)}}{{a^2}{l^2} - 1 + {a^2}{m^2} - 1} ight|$$|{a^2}({l^2} + {m^2}) - 2| = 2\sqrt{{a^4}{l^2}{m^2} - ({a^4}{l^2}{m^2} - {a^2}{l^2} - {a^2}{m^2} + 1)}$$|{a^2}({l^2} + {m^2}) - 2| = 2\sqrt{{a^2}({l^2} + {m^2}) - 1}$Squaring both sides,we get:$[{a^2}({l^2} + {m^2}) - 2]^2 = 4[{a^2}({l^2} + {m^2}) - 1]$

If the portion of the line $lx + my = 1$ falling inside the circle ${x^2} + {y^2} = {a^2}$ subtends an angle of $45^\circ$ at the origin,then

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