Find equation that denotes straight line that intersects middle of chord that is expressed by another equation

Question

$(x-2)^2 + (y + 1)^2 = 16$ is circle

$x - 2y - 3 = 0$ is equation that denotes chord of this circle

How do I find equation of straight line that intersects middle of this chord?

Answer 1

Line $L_2$ perpendicular to $L_1 : x-2y-3=0$ at the midpoint of the chord passes through the center $(2, -1) $.

$$ L_2 : 2x+y=2(2) +(-1) =3$$

Midpoint is their intersection : $(9/5, -3/5) $

Required equation is $y+3/5=m(x-9/5) $

Answer 2

By intersecting, I suppose that you are looking for a line that it is a perpendicular bisector to the given chord (intersecting lines through the mid point are infinite, so I am assuming that you are referring to the line passing through the mid point and perpendicular to the given chord).

If that is the case, then you may proceed as follows ;

1. The perpendicular bisector will have a slope 'm', where m satisfies m*(1/2)=-1 (condition for lines with perpendicular slopes)
2. To find the mid point of the chord, find the coordinates of the foot of perpendicular of the center of the circle (2,-1) in the given chord ( using the foot of perpendicular formula )
3. Using the slope obtained in 1, and the point obtained in 2, one may easily write the equation of the required line.

I am also attaching a picture of the solution for the problem.