Let the equation of the circle be $x^2+y^2+Dx+Ey+F=0$
$(1,2) \implies 1^2+2^2+D.1+E.2+F=0 \implies D+2E+F = -5$ ......$eq(i)$
$(3,-4) \implies 3^2+(-4)^2+D.3+E.(-4)+F=0 \implies 3D-4E+F = -25$ .....$eq(ii)$
$(5,6) \implies 5^2+(6)^2+D.5+E.6+F=0 \implies 5D+6E+F =- 61$.....$eq(iii)$
$eq(i) - eq(ii) \implies -2D +6E = 20$ .....$eq(iv)$
$eq(iii) - eq(ii) \implies 2D +10E = -36$ .....$eq(v)$
$eq(iv) + eq(v) \implies 16E = -16 \implies E=-1$ .....$eq(iv)$
Putting value of $E$ in $eq(iv) \implies -2D = 26 \implies D = -13$
Putting value of $E$ and $D$ in $eq(i) \implies -13 -2 +F = -5 \implies F = 10$
$\therefore D= - 13 , E = - 1, F= 10$
So the equation of circle becomes $x^2+y^2-13x-y+10=0$
Since the line is creating a chord on the circle so it must be intersecting at Two points on the circle and both these points would satisfy the equation of circle as well as equation of given line.i.e.
$x^2+y^2-13x-y+10=0$.... $eq(vi)$ and $3x-4y+5=0$ ....$eq(vii)$
Calculating $x$ from $eq(vii)$ and putting it in $eq(vi)$
$3x-4y+5=0 \implies x = \frac{4y-5}{3}$ ....$eq(viii)$
$ (\frac{4y-5}{3})^2 + y^2 -13*(\frac{4y-5}{3}) -y +10 =0$
$\implies \frac{16y^2+25-40y}{9} + y^2 -13*(\frac{4y-5}{3}) -y +10 =0$
$\implies16y^2+25-40y + 9y^2 -39(4y-5) -9y +90 =0$
$\implies16y^2+25-40y + 9y^2 -156y+195 -9y +90 =0$
$\implies16y^2+ 9y^2 -40y -156y-9y+195 +90+25 =0$
$\implies 25y^2 -205y+310 =0$
$\implies y= \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{205\pm \sqrt{42025-31000}}{50} = \frac{205\pm \sqrt{11025}}{50} = \frac{205\pm 105}{50} = 6.2,2$
Putting values of $y$ in $eq(viii)$
$y=6.2 \implies x = \frac{4*6.2-5}{3} = \frac{19.8}{3} = 6.6$
$y=2 \implies x = \frac{4*2-5}{3} \implies x=1$
So the points of intersection are $(6.6,6.2)$ and $(1,2)$
and the distance between these $2$ points = Length of the chord $= \sqrt{(6.6-1)^2 + (6.2-2)^2} = \sqrt{(5.5)^2 + (4.2)^2 } = \sqrt{47.89} = 6.92$
So NONE OF THE OPTIONS are CORRECT.
