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A family of circles passing through the ...

A family of circles passing through the points `(3, 7) and (6, 5)` cut the circle `x^2 + y^2 - 4x-6y-3=0`. Show that the lines joining the intersection points pass through a fixed point and find the coordinates of the point.

Text Solution

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The correct Answer is:
`x = 2 and y = 23//3`
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
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