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If the radical axis of the circles `x^2+y^2+2gx+2fy+c=0`and `2x^2+2y^2+3x+8y+2c=0`touches the circle `x^2+y^2+2x+1=0`, show that either `g=3/4`or `f=2`A. `g=(4)/(3)` and f=2B. `g=(4)/(3)` and `f=(1)/(2)`C. `g=-(3)/(4)` and `f=2`D. `g=(3)/(4)` and `f=(1)/(2)` |
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Answer» Correct Answer - B Equations of the two circles are `x^(2)+y^(2)+2gx+2fy+c=0 and x^(2)+y^(2)+(3)/(2)x+y+c=0` The equation of radical axis is `(2g-(3)/(2))x+(2f-1)y=0` ...(i) It touches the circle `x^(2)+y^(2)+2x+2y+1=0` `:. |(-(2g-(3)/(2))-(2f-1))/(sqrt((2g-(3)/(2))^(2)+(2f-1)^(2)))|=1` `rArr(2g+2f-(5)/(2))^(2)=(2g-(3)/(2))^(2)+(2f-1)^(2)` Clearly `g=(4)/(3) and f=(1)/(2)` satisfy this relation. |
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