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. (a) ( 1 / 2 , − 5 / 2 ) (b) ( 1 3 , 4 / 3 ) (c) ( − 1 / 2 , 3 / 2 ) (d) none of these

Options

A. (a) `(1//2,-5//2)`
B. (b) `(1/3,4//3)`
C. (c) `(-1//2,3//2)`
D. (d) none of these

Answer

1

Explanation

Let the equation of the circle be
`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.