Consider a family of circle passing thorugh two fixed points A(3,7) and B(6,5) . Show that the chords in which the circle `x^(2)+y^(2)-4x-6y-3=0`. Cuts the members of the family are concurrent at a point. Find the coordinates of this point.

Answer

`x = 2 and y = 23//3`

Explanation

The equation of the circle on the line joining the points A (3, 7) and B (6, 5) as diameter is
`(x-3)(x-6)+(y-7)(y-5)=0" "...(i)`
and the equation of the line joining the point A (3, 7) and B (6, 5) is `y-7=(7-6)/(3-6)(x-3)`
`rArr 2x + 3y - 27 = 0" "...(ii)`
Now, the equation of family of circles passing through the point of intersection of Eqs. (i) and (ii) is
`S+lambdaP=0`
`rArr (x-3)( x-6)+(y-7)(y-5)+lambda(2x+3y-27)=0`
`rArr x^(2)-6x-3x+18+y^(2)-5-7y+35`
`+2lambdax+3lambday-27lambda=0`
`rArr S_(1)-=x^(2)+y^(2)+x(2lambda-9)+y(3lambda-12)`
`+(53-27lambda)=0" "...(iii)`
Again, the circle,which cut the members of family of circles, is
`S_(2)-=x^(2)+y^(2)-4x-6y-3=0" "...(iv)`
and the equation of common chord to circles `S_(1) and S_(2)` is
`S_(1)-S_(2)=0`
`rArrx(2lambda-9+4)+y(3lambda-12+6)+(53-27lambda+3)=0`
`rArr 2lambdax - 5x +3lambday-6y+56-27lambda=0`
`rArr (-5x-6y+56)+lambda(2x+3y-27)=0`
which represents equations of two straight lines passing through the fixed point whose coordinates are obtained by solving the two equations
5x + 6y - 56 = 0 and 2x + 3y - 27 = 0
we get x = 2 and y = 23/3