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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. |
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Answer» Correct Answer - `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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