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Consider the relation 4l^(2)-5m^(2)+6l+1=0 , where l,m in R Tangents PA and PB are drawn to the above fixed circle from the points P on the line x+y-1=0 . Then the chord of contact AP passes through the fixed point. |
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Answer» `(1//2,-5//2)` `X^(2)+y^(2)+2gx+2fy+c=0`(1) The line `lx +my+1=0` will touch circle (1) if the length of perpendicular from the center `( -g, -f)` of the circle on the line is equal to its radius, i.e., `(|-g l =mf +1|)/(sqrt(l^(2)+m^(2)))=sqrt(g^(2)+f^(2)-c)` `(gl+mf-1)^(2)= (l^(2)+m^(2))(g^(2)+f^(2)-c)` or `(c-f^(2))l^(2)+(c-g^(2))m^(2)-2gl-2fm+2fglm+1=0`(2) But the given condition is `4l^(2)-5m^(2)+6l+1=0`(3) Comparing (ii) and (iii), we GET `c-f^(2)=4,c-g^(2)= -5, -2g = 6, -2f =0, 2gf=0` Solving, we get `f=0, g= -3, c=4` Substituting these values in (1) , the equation of the circle is `x^(2)+y^(2)-6x+4=0`. Any point on the ling `x+y-1=0` is `(t, 1-t) , t in R`. The chord of contact w.r.t. this point of circle is `TX +y(1-t) -3(t+x) +4 =0` or `t ( x-y-3) + (-3x+y+4)=0`, which is concurrent at the point of intersection of the lines `x-y-3=0` and `-3x+y+4=0` for all values of t. Hence, the lines are concurrent at `(1//2, -5//2)`. Also point (2,-3) lies outside the circle from which TWO tangents can be drawn. |
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