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If `O Aa n dO B`are equal perpendicular chordsof the circles `x^2+y^2-2x+4y=0`, then the equations of `O Aa n dO B`are, where `O`is the origin.`3x+y=0`and `3x-y=0``3x+y=0`and `3y-x=0``x+3y=0`and `y-3x=0``x+y=0`and `x-y=0`A. `3x+y=0` and `3x-y=0`B. `3x+y=0` and `3y-x=0`C. `x+3y=0` and `y-3x=0`D. `x+y=0` and `x-y=0` |
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Answer» Correct Answer - 3 Let the equation of the chord OA of the circle `x^(2)+y^(2)-2x+4y=0` (1) by `y=mx` (2) Solving (1) and (2) , we get `x^(2)+m^(2)x^(2)-2x+4x=0` or `(1+m^(2))x^(2)-(2-4m)x=0` or `x=0` and `x=(2-4m)/(1+m^(2))` Hence, the points of intersection are `O(0,0)` and `A((2-4m)/(1+m^(2)),(m(2-4m))/(1+m^(2)))` or `OA^(2)=((2-4m)/(1+m^(2)))(1+m^(2))=((2-4m)^(2))/(1+m^(2))` Since OAB is an isosceles right-angles triagnle, `OA^(2)=(1)/(2)AB^(2)` where AB is a diameter of the given circle. Hence, `OA^(2)=10` or `((2-4m)^(2))/(1+m^(2))=10` or `4-16m+16m^(2)=10+10m^(2)` or `3m^(2)-8m-3=0` i.e., `m=3` or `-(1)/(3)` Hence, the required equations are `y=3x` or `x+3y=0` |
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