It is given that the straight line $l : y = mx + c$ intersects the circle $C : x^2 + y^2 = r^2$

Question

It is given that the straight line $l : y = mx + c$ intersects the circle $C : x^2 + y^2 = r^2$ at points $A$ and $B$. Let $M$ be the midpoint of $AB$ and $l'$ be the perpendicular bisector of $AB$. Using an algebraic method, show that $l'$ passes through the centre $O$, i.e., perpendicular bisector of a chord passes through the centre of the circle.

Accepted Answer

$l':x+my=0$, so $O=(0,0)$ lies on the perpendicular bisector (including $m=0$, when $l':x=0$).

Coordinate Geometry - Line–Circle Intersections

Coordinate Geometry - Line–Circle Intersections

Solution:

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