SSC CGL 2024 Tier-I · 10 September 2024, Shift 1 · Quantitative Aptitude · Question 5 of 98

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

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SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

X, Y, and Z are three points on a plane with XY = 11 cm, YZ = 13 cm, and XZ = 24 cm. The number of circles that travel through points X, Y, and Z is:

(a)0
(b)2
(c)1
(d)3

Answer

Why

Correct — A. Test the triangle inequality before reaching for any circle formula.

XY + YZ = 11 + 13 = 24 cm = XZ

That is equality, not the strict inequality a triangle needs. So X, Y and Z do not enclose a triangle: Y sits on the segment XZ, and the three points are collinear.

A circle through three points is centred where the perpendicular bisectors of the joining segments meet. For collinear points those bisectors are all perpendicular to the same line, so they are parallel and never meet.

No centre exists, so no circle passes through all three — option (a).

Why the others are wrong

(b)
2 — No set of three distinct points ever admits exactly two circles. Three non-collinear points give exactly one, and three collinear points give none.
(c)
1 — One is the count for three non-collinear points, whose perpendicular bisectors meet at the circumcentre. Here 11 + 13 = 24 makes the points collinear and that construction fails.
(d)
3 — Three circles would need three different points each equidistant from X, Y and Z, and no such point exists at all once the three lie on one line.

Concept

Through any three non-collinear points there passes exactly one circle. Its centre is the circumcentre, where the perpendicular bisectors of the joining segments meet, and its radius is the distance from that centre to any of the three.

Collinearity destroys the construction. The perpendicular bisectors of collinear segments are all perpendicular to the same line, hence parallel, so no point is equidistant from all three and no circle can contain them.

Comparing the two shorter distances against the longest is the quickest collinearity test a question like this leaves you.

The lengths are chosen so that 11 + 13 lands exactly on 24. One centimetre either way and the answer becomes 1, which is why the sum has to be computed rather than eyeballed.

Key facts

Exactly one circle passes through three non-collinear points.
No circle passes through three collinear points.
Three points are collinear when the two shorter distances add to the longest.
The circumcentre is the intersection of the perpendicular bisectors of the sides.

Study next

Circumcentre and circumradius of a triangleThe triangle inequality as a test on three given lengthsHow many circles pass through one point, two points and three points

Common traps

Assuming three points always give exactly one circle and answering 1 without adding 11 and 13.
Reading XY + YZ = XZ as a thin triangle rather than as outright collinearity.

The same fact is asked head-on at 25 Sep 2024, 09:00, Quant Q.12 — it is always possible to draw a circle through ______ non-collinear points in a plane, keyed to 3. Here the word non-collinear is removed and hidden behind a length check instead.

Related PYQs

No directly related past PYQ was found.