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The equation of the circle whose diameter is the common chord of thecircles; `x^2+y^2+3x+2y+1=0`& `x^2+y^2+3x+4y+2=0`is:`x^2+y^2+8x+10 y+2=0``x^2+y^2-5x+4y+7=0``2x^2+2y^2+6x+2y+1=0`None of theseA. `x^(2)+y^(2)+8x+10y+2=0`B. `x^(2)+y^(2)-5x+4y+7=0`C. `2x^(2)+2y^(2)+6x+2y+1=0`D. none of these |
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Answer» Correct Answer - C The equation of the common chord of the circles `x^(2)+y^(2)+3x+2y+1=0 and x^(2)+y^(2)+3x+4y+2=0` is given by 2y+1=0 [Using : `S_(1)-S_(2)=0`] The equation of a circle passing through the intersection of the given circles is `(x^(2)+y^(2)+3x+2y+1)+lambda(x^(2)+y^(2)+3x+4y+2)=0` `rArr x^(2)+y^(2)+x((3+3lambda)/(lambda+1))+2lambda((1+2lambda)/(lambda+1))+(1+2lambda)/(lambda+1)=0` ...(i) Since 2 y+1=0 is a diameter of this circle. Therefore, its centre `(-(3)/(2), -(2lambda+1)/(lambda + 1))` lies on it. `:. - 2 ((2lambda+1)/(lambda+1))+1=0rArr - 4 lambda-2 + lambda+1=0rArr lambda=-(1)/(3)` Putting `lambda=-1//3` in (i), the equation of the required circle is `2x^(2)+2y^(2)+6x+2y+1=0` |
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