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Consider a family of circles passing through the point (3,7) and (6,5). Answer the following questions. If the circle which belongs to the given family cuts the circle x^(2)+y^(20=29 orthogonally, then the center of that circle is |
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Answer» `(1//2,3//2)` `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) If the circle BELONGING to this family touches the x-axis , then equation `(x-3)(x-6)+(0-7)(-5)+lambda { 2x+3(0)-27}=0` has two equal roots, for which discriminanat `D=0`. It gives two values of `lambda`. The equation of the common chord 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)x+(3 lambda -6)y +(-27 lambda + 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)`. If circle `(i)` CUTS `x^(2)+y^(2)=29` orthogonally , then `0+0= -29 +56-27 lambda =0` or `LAMDA =1` So, the required circle is `x^(2)+y^(2)-7x-9y+26=0` and the center is `(7//2,9//2)`. |
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