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consider a family of circles passing through two fixed points `S(3,7)` and `B(6,5)`. If the common chords of the circle `x^(2)+y^(2)-4x-6y-3=0` and the members of the family of circles pass through a fixed point (a,b), then |
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Answer» The equation of the line passing throught the points A(3,7) and B(6,5) is `y-7=-(2)/(3)(x-3)` or `2x+3y-27=0` Also, the equation of the circle with A and B as the endpoints of diameter is `(x-3)(x-6)+(y-7)(y-5)=0` Now, the equation of the family of circles through A and B is `(x-3)(x-6)+(y-7)(y-5)+lambda(2x+3y-27)=0` (1) The equation of the common chrod of (1) and `x^(2)+y^(2)-4x-6y-3=0` is the radical axis, which is `[(x-3)(x-6)+(y-7)(y-5)+lambda(2x+3y-27)]-[x^(2)+y^(2)-4x-6y-3]=0` or `(2 lambda-5)+(3lambda-6)y+(-27lambda+56)=0` or `(-5x-6y+56)+lambda(2x+3y-27)=0` This is the family of lines which passes through the point of intersection of `-5x-6y+56=0` and `2x+3y-27=0`,` i.e., `(2, 23//3)`. |
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