Parts of a circle

Draw a circle and you can already point to most of its pieces from memory: a centre, a radius reaching out to the edge, maybe a chord cutting across. What turns this vocabulary into real problem-solving power is one extra fact that ICSE Class 9 geometry adds — drop a perpendicular from the centre onto any chord, and it always lands exactly at the chord's midpoint. That single sentence, paired with the Pythagoras theorem you already know, is enough to compute chord lengths, distances from the centre, and radii that would otherwise need much heavier tools. This article names every part of a circle carefully, states the perpendicular-bisects-chord fact, and then puts it to work in a fully solved numeric example.

The core insight

The perpendicular from a circle's centre to a chord bisects the chord, turning the radius, that perpendicular distance, and half the chord into a right-angled triangle solvable by the Pythagoras theorem.

The core idea

The vocabulary. Every circle is built from the same named parts, and being fluent in them is half the battle in any circle problem.

  • The centre is the one fixed point every point on the circle is equally far from.
  • The radius (rr) is the distance from the centre to any point on the circle.
  • The diameter (dd) is a chord that passes through the centre — the longest possible chord — and is always twice the radius.
  • A chord is any straight segment joining two points on the circle; a diameter is just the special chord that happens to pass through the centre.
  • An arc is a curved piece of the circle's boundary between two points (the shorter one is the minor arc, the longer one the major arc); a sector is the pie-slice region between two radii and their arc; a segment is the region between a chord and the arc it cuts off.

The diameter-radius and circumference relationships are simple multiples of rr:

d=2r,C=2πr=πdd = 2r, \qquad C = 2\pi r = \pi d

where: rr is the radius, dd the diameter, and CC the circumference — the total distance around the circle.

The chord-bisection fact. If you drop a perpendicular from the centre OO onto a chord, it lands exactly at the chord's midpoint — it bisects the chord. This builds a right-angled triangle with the radius as the hypotenuse, the perpendicular distance from centre to chord as one leg, and half the chord as the other leg, so the Pythagoras theorem applies directly:

r2=(chord2)2+(distance from centre)2r^2 = \left(\frac{\text{chord}}{2}\right)^2 + (\text{distance from centre})^2

where: rr is the radius, chord is the full chord length, and distance from centre is the perpendicular distance from OO to that chord. Given any two of these three quantities, the third follows immediately.

Figure
A circle with centre O a chord AB a perpendicular from O meeting AB at its midpoint M and the right triangle formed by radius OA perpendicular OM and half chord AM
Perpendicular from centre bisects the chord O A B M radius r distance
Derivation AChord length or distance from the Pythagoras theorem

Given a circle of known radius and a chord at a known perpendicular distance from the centre, the goal is to find the length of that chord — and, run the other way, given the radius and the chord length, to find how far that chord sits from the centre — using only the fact that the perpendicular from the centre bisects the chord and the ordinary Pythagoras theorem, with no new circle theory beyond that one bisection fact.

Step 1. Set up the right triangle. Draw the radius OAOA to one endpoint of the chord, and drop the perpendicular OMOM from the centre OO onto the chord. Because the perpendicular bisects the chord, MM is the chord's midpoint, so triangle OMAOMA is right-angled at MM, with hypotenuse OA=rOA = r, one leg OM=OM = the distance from centre to chord, and the other leg MA=MA = half the chord.

Step 2. Write the Pythagoras relation for this triangle. OA2=OM2+MA2OA^2 = OM^2 + MA^2, which is exactly r2=(distance)2+(chord2)2r^2 = (\text{distance})^2 + \left(\dfrac{\text{chord}}{2}\right)^2. This one equation connects all three quantities, so knowing any two pins down the third.

Step 3. Solve for whichever quantity is unknown. To find the chord, rearrange to (chord2)2=r2−(distance)2\left(\dfrac{\text{chord}}{2}\right)^2 = r^2 - (\text{distance})^2, take the square root, then double the result. To find the distance instead, rearrange to (distance)2=r2−(chord2)2(\text{distance})^2 = r^2 - \left(\dfrac{\text{chord}}{2}\right)^2 and take the square root directly — no doubling needed since distance is not halved in the original figure.

Step 4. Sanity-check the answer. The distance from the centre can never exceed the radius, and half the chord can never exceed the radius either; if a computed value comes out larger than rr or a square root turns negative, a number has been misread or swapped.

Worked example

A circle has radius 1010 cm. A chord is drawn so that its perpendicular distance from the centre is 66 cm. Find the length of the chord, and then check the answer against a second circle of diameter 2626 cm carrying a 2424 cm chord.

Step 1. Identify the right triangle for the first circle. The radius 1010 cm is the hypotenuse, the distance from centre to chord (66 cm) is one leg, and half the chord is the other leg.

Step 2. Apply the Pythagoras theorem. (chord2)2=r2−(distance)2=102−62=100−36=64\left(\dfrac{\text{chord}}{2}\right)^2 = r^2 - (\text{distance})^2 = 10^2 - 6^2 = 100 - 36 = 64.

Step 3. Solve for the chord. chord2=64=8\dfrac{\text{chord}}{2} = \sqrt{64} = 8, so chord =2×8=16= 2 \times 8 = 16 cm.

Step 4. Check the reverse direction on the second circle. Diameter 2626 cm gives radius r=13r = 13 cm, and a 2424 cm chord has half-chord 1212 cm, so (distance)2=132−122=169−144=25(\text{distance})^2 = 13^2 - 12^2 = 169 - 144 = 25, giving distance =25=5= \sqrt{25} = 5 cm — confirming the same relation works equally well solving for the distance.

The chord in the first circle is 1616 cm long, and the 2424 cm chord in the second circle sits 55 cm from the centre.

Common traps

  • Confusing radius and diameter. The diameter is exactly twice the radius and always passes through the centre; forgetting to halve or double between the two is a very common slip.
  • Assuming every chord is a diameter. A diameter is only the longest possible chord — most chords do not pass through the centre at all.
  • Mixing up arc and chord. An arc is the curved edge of the circle between two points; a chord is the straight segment joining the same two points. Only the chord can be used with the Pythagoras theorem above.
  • Forgetting to halve the chord before applying Pythagoras. It is half the chord, not the full chord, that forms the right-angled triangle with the radius and the distance from the centre.

Ask the tutor: if the right-triangle picture still feels shaky, ask your tutor to draw one circle, mark the centre, drop the perpendicular to a chord, and shade the little right triangle formed by the radius, the perpendicular, and half the chord.

Why does the perpendicular from the centre bisect the chord instead of just crossing it somewhere?

Picture the two radii drawn to the chord's endpoints — they are equal in length since both are radii of the same circle, so the triangle they form with the chord is isosceles. The perpendicular from the centre is also the triangle's line of symmetry, and in an isosceles triangle the line of symmetry from the apex always meets the base at its exact midpoint. That symmetry is the real reason the bisection happens every single time, not just in convenient examples.

Do I always need to know the radius to use the Pythagoras relation on a chord?

You need any two of the three quantities — radius, chord length, and distance from the centre — to find the third; it does not have to be the radius that is missing. If you know the chord and the distance from the centre, you can find the radius; if you know the radius and the chord, you can find the distance, exactly as in the worked example's second check.

Is a diameter just a special chord, or a completely different thing?

It is a special chord, not a different object. Every diameter satisfies the definition of a chord — a straight segment joining two points on the circle — it just happens to also pass through the centre, which makes it the longest chord possible. If you ever apply the Pythagoras relation to a diameter itself, the distance from the centre is zero, since the diameter passes right through OO.

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