about to only mathematics

Options

A. 0
B. 1
C. 2
D. infinite

Answer

3

Explanation

The equation of the line passing through 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)
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)`.