Chapter 10 · Kerala SSLC Class 10 Maths
Circles And Lines
Two ways to learn — a crisp formula reference, or a story that makes the relationships feel obvious.
Four scenes that build the relationships from scratch, step by step.
How this works
Each scene turns a circle-and-line problem into one clean relationship. Read the story, then tap "The Maths" to see the formal result.
Two Wires, One Crossing
A circular tabletop has two thin wires stretched across it, from rim to rim, crossing each other somewhere in the middle — not at the centre. Call the wires AB and CD, and their crossing point P.
I measured AP as 4 cm and PB as 6 cm on the first wire. On the second, CP is 3 cm. What's PD, without me having to measure it directly?
Join AC and BD to make two triangles, APC and DPB. The angle at P is the same in both (vertically opposite angles). And the angle at A equals the angle at D — both are angles standing on the same arc BC, looked at from opposite sides of the circle.
The sides opposite the matching angles pair up as (AP, DP) and (CP, BP). If both pairs scale by the same factor k, then AP = k·DP and CP = k·BP is one way to write it — rearranged, it gives a product rule that doesn't need k at all.
AP × PB = CP × PD
4 × 6 = 3 × PD
24 = 3 × PD
PD = 8 cm
The Broken Ring
A piece of an old metal ring is dug up — just an arc, the rest long gone. An archaeologist wants to know the ring's original diameter, using only what remains.
If I draw the diameter through this arc and drop a perpendicular chord to it, I can measure where they cross. One part of the diameter measures 2 cm from the edge. The chord itself splits into two equal halves of 3 cm each, since the diameter is perpendicular to it.
This is the crossing-chords idea again — but now one of the two chords is a diameter, and the other is perpendicular to it. That special combination bisects the second chord automatically, splitting it into two equal halves.
a × b = c² (a, b = the two parts of the diameter; c = half the chord)
2 × (d − 2) = 3 × 3 = 9
d − 2 = 4.5
d = 6.5 cm
This idea — a product of two lengths turning into a perfect square — is exactly how ancient mathematicians like Bhaskaracharya constructed lengths involving square roots using nothing but a circle and a straightedge, long before algebraic notation existed.
The Point Inside
Zoom out from any one chord. Fix a single point P inside a circle, at a known distance d from the centre. Now draw many different chords through P — as many as you like, in every direction.
Every chord through P gets cut into two different-looking parts. Do all these different products still come out equal?
Yes — and there's a clean way to compute that common product directly from the circle's radius r and the point's distance d from the centre, without measuring any chord at all. Draw the diameter through P: it's cut into parts (r + d) and (r − d).
PX × PY = (r + d)(r − d)
= r² − d²
The Point Outside
Now step outside the circle. From an external point P, draw two different secant lines, each cutting the circle at two points.
Does the same rule still hold, even though the lines don't cross INSIDE the circle this time?
Yes — the same similar-triangle argument from Scene 1 still works here, just with the point outside instead of inside. And the same distance-based shortcut works too, just with r and d swapped:
PA × PB = PC × PD
= (d + r)(d − r)
= d² − r²
Now push one of the secants until it just grazes the circle at a single point — it becomes a tangent, of length PT. The two intersection points merge into one, so PX × PY collapses to PT × PT.
The Big Picture
One idea runs through this whole chapter: a point's "power" with respect to a circle — a single number depending only on the radius and the point's distance from the centre — controls every chord, secant, or tangent drawn through it.
Two chords crossing inside?
AP × PB = CP × PD
Diameter cut by a perpendicular chord?
a × b = c²
Point inside the circle?
PX × PY = r² − d²
Point outside the circle?
PX × PY = d² − r²
One line is a tangent?
PX × PY = PT²
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