EPFO APFC 2023 General Studies · Question 73 of 120

Correct — D, (d) √13/2, printed in the booklet as the square root of 13 over 2. The quickest route uses a circle theorem rather than algebra. The three points are the origin, a point on the y-axis and a point on the x-axis, so the segment from the origin to (0, 3) and the segment from the origin to (2, 0) are perpendicular, and the angle subtended at the origin by the chord joining (0, 3) and (2, 0) is a right angle. An angle in a semicircle is a right angle, and the converse holds as well: if a chord subtends a right angle at a point on the circle, that chord is a diameter. The distance from (0, 3) to (2, 0) is the square root of 2 squared plus 3 squared, which is the square root of 13, so the diameter is the square root of 13 and the radius is half of it, √13/2, roughly 1·80. The algebraic route gives the same thing and is worth knowing because it works when no right angle is available. Take the circle as x squared plus y squared plus 2gx plus 2fy plus c equals 0. Substituting the origin gives c equal to 0. Substituting (2, 0) gives 4 plus 4g equal to 0, so g is minus 1. Substituting (0, 3) gives 9 plus 6f equal to 0, so f is minus three halves. The radius is the square root of g squared plus f squared minus c, which is the square root of 1 plus nine quarters, that is the square root of thirteen quarters, which is √13/2. A third route is purely geometric: the centre is equidistant from the origin and (0, 3), so it lies on the line y equals three halves, and equidistant from the origin and (2, 0), so it lies on the line x equals 1. The centre is therefore (1, 3/2) and its distance from the origin is again √13/2. Three non-collinear points determine exactly one circle, so all three methods must agree.

• (a)2√3 — This value is about 3·46, which is nearly twice the true radius of about 1·80, and it is also the square root of 12, sitting confusingly close to the square root of 13 that this problem produces. The square root of 13 is the diameter of the circle, not its radius, so a candidate who computes the distance from (0, 3) to (2, 0) correctly and then forgets that the chord is a diameter rather than a radius is looking for a number near 3·6, and this is the only option in that neighbourhood. It is not equal to it, and the difference is the point: getting to the square root of 13 is most of the work, and halving it is the step the option is designed to catch. Note also the shape of the two option pairs on the page, one written as a whole number times a surd and the other as a surd over 2. That difference in form is itself a hint that halving is what separates them.
• (b)2√5 — This is the square root of 20, about 4·47, which is two and a half times the correct radius and would give a circle of diameter almost 9 passing through three points that all lie within two units of the y-axis and three units of the x-axis. It can be discarded on scale alone, without any computation, provided the candidate has formed a rough mental picture of the configuration: a circle through the origin, a point three units up and a point two units across cannot be that large. Sketching the three points before starting is worth the few seconds it costs on any coordinate geometry item, because it converts several of the options into obvious impossibilities and gives a check on the answer that is independent of the algebra used to reach it.
• (c)√11/2 — This is the near-miss, differing from the answer only in the number under the radical, and it is what a particular arithmetic slip produces. In the algebraic method the radius squared is g squared plus f squared minus c, which here is 1 plus nine quarters. A candidate converting the 1 into quarters and writing it as two quarters instead of four quarters gets eleven quarters rather than thirteen quarters, and the square root of that is √11/2, about 1·66 against the true 1·80. The two values are close enough that neither a sketch nor a rough estimate will separate them, so this option can only be eliminated by doing the fraction correctly. The general safeguard is to check any answer of this form against a second method; the right-angle argument gives the diameter as the square root of 13 directly, with no fractions to convert at all.

Three non-collinear points determine exactly one circle, and there are three standard ways of finding it. The general-equation method writes the circle as x squared plus y squared plus 2gx plus 2fy plus c equals 0, substitutes each point to get three linear equations in g, f and c, and then uses the centre at minus g, minus f with radius the square root of g squared plus f squared minus c. The perpendicular-bisector method uses the fact that the centre is equidistant from any two points of the circle and so lies on the perpendicular bisector of the chord joining them; two such bisectors intersect at the centre. The circle-theorem method applies when the configuration is special, as it is here: the angle in a semicircle is a right angle, and conversely a chord subtending a right angle at a point of the circle is a diameter. That last theorem, usually learnt as Thales' theorem, converts this problem into a single application of the distance formula. Points placed on the axes, as they are in this question, are the standard signal that the right-angle route is available, because two segments drawn from the origin along the axes are automatically perpendicular. Knowing all three methods matters less than knowing which is cheapest for the configuration in front of you, and recognising the right angle here reduces a three-equation problem to one subtraction and one square root.

Coordinate geometry in this paper is set at a level where the formulae are standard and the discrimination comes from choosing an efficient route and executing small arithmetic cleanly. That is why the option set contains a value differing from the answer only under the radical: the item is testing fraction arithmetic as much as geometry. Two habits pay for themselves here. The first is to sketch the configuration before computing, because a rough picture disposes of options that are of the wrong order of magnitude, and on this question it eliminates half the set in a few seconds. The second is to look for structure before reaching for a general method — points on the axes, right angles, symmetry about a line, integer coordinates — since a special configuration usually has a one-line solution while the general method has four or five lines in which to make a slip. Note also how the options are printed: two of them as a whole number multiplying a surd and two as a surd divided by 2, which the booklet sets as stacked fractions with the radical extending over the number. That visual pairing tells you the setter expects the halving step to be the point of failure, and it is worth pausing over the final division whenever the option list is arranged that way.

• Three non-collinear points determine a unique circle, so all correct methods must yield the same radius; for the points (0, 0), (0, 3) and (2, 0) that radius is √13/2, about 1·80, with centre at (1, 3/2).
• The converse of the angle-in-a-semicircle theorem states that a chord subtending a right angle at a point on the circle is a diameter; here the chord from (0, 3) to (2, 0) subtends a right angle at the origin, so it is a diameter of length √13.
• In the general form x squared plus y squared plus 2gx plus 2fy plus c equals 0, the centre is at minus g, minus f and the radius is the square root of g squared plus f squared minus c; a circle through the origin has c equal to 0.
• The perpendicular-bisector method places the centre on x equals 1, from the chord between (0, 0) and (2, 0), and on y equals three halves, from the chord between (0, 0) and (0, 3), fixing it at (1, 3/2).
• The distance between (0, 3) and (2, 0) is the square root of 2 squared plus 3 squared, that is the square root of 13, and halving it is the step this question is built to test.

• Computing the distance between the two non-origin points, which is the square root of 13, and reporting it as the radius when it is the diameter
• Converting 1 into quarters as two quarters instead of four quarters when evaluating 1 plus nine quarters, which turns thirteen quarters into eleven quarters and produces the near-miss option
• Forgetting that a circle through the origin has no constant term, so that c equals 0 and the radius reduces to the square root of g squared plus f squared
• Reaching for the general three-equation method when the points lie on the axes and the right-angle route settles the problem in one step
• Skipping the sketch; a rough picture of three points close to the origin makes radii of about 3·5 and 4·5 visibly impossible

The paper's quantitative block sets coordinate geometry as short computational items with clean numbers: the radius or centre of a circle through given points, the distance between two points or from a point to a line, the area of a triangle from its vertices, or the equation of a line through a point with a stated slope or intercept. The configurations are chosen so that a theorem shortens the work, and the option sets are built to catch a specific slip — a factor of two, a sign, or a fraction wrongly converted — rather than a wholesale misunderstanding. The preparation that pays is fluency with the handful of standard formulae plus the habit of scanning for structure first, and the checking discipline that matters most is verifying the final answer by a second, independent route, since these problems are short enough that two methods still cost less than one careless method plus a wrong mark.

No directly related past PYQ was found.

• practice — not a real PYQ

  What is the centre of the circle that passes through the points (0, 0), (0, 3) and (2, 0)?

  • (a)(1, 3/2)
  • (b)(2, 3)
  • (c)(3/2, 1)
  • (d)(0, 0)

  Answer(a) (1, 3/2) — the centre is equidistant from (0, 0) and (2, 0), so it lies on the line x equals 1, and equidistant from (0, 0) and (0, 3), so it lies on the line y equals three halves. It is also the midpoint of the chord joining (0, 3) and (2, 0), which is a diameter because it subtends a right angle at the origin, and the distance from this centre to any of the three points is √13/2.

• practice — not a real PYQ

  A chord of a circle subtends a right angle at a point on the circumference of the same circle. It follows that the chord is

  • (a)a tangent to the circle
  • (b)a diameter of the circle
  • (c)equal in length to the radius
  • (d)perpendicular to the radius at its midpoint only if the circle is a unit circle

  Answer(b) a diameter of the circle — this is the converse of the theorem that an angle in a semicircle is a right angle, and it is the fastest way to find a circle through three points when two of the chords from one of them are perpendicular. It is what turns the problem of a circle through (0, 0), (0, 3) and (2, 0) into a single application of the distance formula, since the chord joining the two axis points is then a diameter.