Multiple choice

$A,B,C$ are three points on a circle such that $AB$ is the chord and $CP$ is the perpendicular to $OP$, where $O$ is the centre and $P$ is any point on $AB$. The radius $r$ of the circle is given by

  1. ${ r }^{ 2 }={ OP }^{ 2 }+AP\times CP$
  2. ${ r }^{ 2 }={ OP }^{ 2 }+AP\times PB$
  3. ${ r }^{ 2 }={ OP }^{ 2 }+PB\times PC$
  4. ${ r }^{ 2 }={ OP }^{ 2 }+{ PB }^{ 2 }$
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B Correct answer
Explanation

Let O be the origin. P is on AB. CP is perpendicular to OP. This is a property of chords in a circle. By the power of a point theorem or geometric construction, the relation r^2 = OP^2 + AP*PB holds.

AI explanation

Draw the radius OA from the center O to point A, forming the right triangle OAP. By the Pythagorean theorem, the radius squared r^2 equals OP^2 plus AP^2. Since the line from the center perpendicular to a chord bisects the chord, extend OP to intersect the circle, applying the intersecting chords theorem where OP multiplied by the remaining segment equals AP multiplied by PB. Substituting this relation into the Pythagorean result gives r^2 = OP^2 + AP x PB.

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