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

math-traveler
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  • There are infinitely many lines passing through the midpoint of chord in all directions. Can you determine the coordinates of midpoint first? – cosmo5 Jan 15 '21 at 08:50
  • Once you've determined the intersections of the line and the circle, proceed to find the midpoint of that chord. It's then easy to find lines (in fact, infinitely many) passing through this midpoint – Dr. Mathva Jan 15 '21 at 08:51
  • Thanks, I got it – math-traveler Jan 15 '21 at 08:53

2 Answers2

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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) $

cosmo5
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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.