<p>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&nbsp;\(M\) be the midpoint of \(AB\) and \(l'\) be the perpendicular bisector of \(AB\).</p>
<p>Using an algebraic method, show that&nbsp;\(l'\) passes through the centre \(O\), i.e., perpendicular bisector of a chord passes through the centre of the circle.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1772868560708-izi65rup6.png" style="max-width: min(300px, 100%); height: auto; display: block;" alt="Diagram showing the intersection of line l and circle C at points A and B, with midpoint M and perpendicular bisector l'" class="alignright custom-width" /></figure></div>

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.

Further Coordinate Geometry

Further Coordinate Geometry