Question 242 marksVery Short AnswerSection B

  1. The coordinates of the centre of a circle are (x−7,2x)(x - 7, 2x). Find the value(s) of 'x', if the circle passes through the point (−9,11)(-9, 11) and has radius 525\sqrt{2} units.

Coordinate Geometry
Distance formula / equation of circle

Official answer

x=3x = 3 or x=5x = 5. Setting the distance from centre (x−7,2x)(x - 7, 2x) to point (−9,11)(-9, 11) equal to the radius 525\sqrt{2} and squaring gives (x+2)2+(2x−11)2=50(x + 2)^2 + (2x - 11)^2 = 50, which simplifies to x2−8x+15=0x^2 - 8x + 15 = 0, factorising as (x−3)(x−5)=0(x - 3)(x - 5) = 0.

Answer keywords:distance formula equals radiuscircle centre and radius(5√2)² = 50x² − 8x + 15 = 0(x−3)(x−5)=0x = 3 or x = 5two values of x

Marking scheme

  1. 1.½ mark: Setting up distance = radius: (x+2)2+(2x−11)2=(52)2=50(x + 2)^2 + (2x - 11)^2 = (5\sqrt{2})^2 = 50.
  2. 2.½ mark: Correct expansion and simplification to 5x2−40x+75=05x^2 - 40x + 75 = 0 (or x2−8x+15=0x^2 - 8x + 15 = 0).
  3. 3.½ mark: Correct factorisation (x−3)(x−5)=0(x - 3)(x - 5) = 0.
  4. 4.½ mark: Both values x=3x = 3 and x=5x = 5 stated (dropping one root loses this half-mark).

Hint

Distance from the centre to the given point equals the radius. Write CP2=(52)2=50CP^2 = (5\sqrt{2})^2 = 50, and remember (x−7)−(−9)=x+2(x - 7) - (-9) = x + 2. Expect TWO values of x.

Quick oral answer

Since the point (−9,11)(-9, 11) lies on the circle, its distance from the centre (x−7,2x)(x - 7, 2x) equals the radius 525\sqrt{2}; squaring gives (x+2)2+(2x−11)2=50(x + 2)^2 + (2x - 11)^2 = 50, which simplifies to x2−8x+15=0x^2 - 8x + 15 = 0, so x=3x = 3 or x=5x = 5, both giving valid centres.

Common mistakes

  1. 1.Sign error in (x−7)−(−9)(x - 7) - (-9): writing (x−16)(x - 16) instead of (x+2)(x + 2) by mishandling the double negative.
  2. 2.Forgetting to square the radius correctly — using (52)2=10(5\sqrt{2})^2 = 10 or 525\sqrt{2} instead of 50.
  3. 3.Reporting only one root (x=3 OR x=5)(x = 3 \text{ OR } x = 5) when the equation clearly yields both, ignoring the plural 'value(s)' in the question.

Analysis and explanation

This question links the distance formula to the geometric definition of a circle (every point on it is equidistant from the centre).


Concept

  • Point P(−9,11)P(-9, 11) lies on the circle with centre C(x−7,2x)C(x - 7, 2x), so distance CP equals the radius 525\sqrt{2}.
  • Squaring removes the surd immediately: (52)2=50(5\sqrt{2})^2 = 50.

Key points

  • Setting up (x+2)2+(2x−11)2=50(x + 2)^2 + (2x - 11)^2 = 50 and simplifying gives the clean quadratic x2−8x+15=0x^2 - 8x + 15 = 0, which factorises as (x−3)(x−5)=0(x - 3)(x - 5) = 0.
  • Both roots x=3x = 3 and x=5x = 5 are valid — the phrase 'value(s)' signals more than one answer is expected.

Common mistakes

  • Sign slip in (x−7)−(−9)=x+2(x - 7) - (-9) = x + 2 (double negative).
  • Discarding one of the two valid roots instead of reporting both.

Interesting facts

Two distinct centres satisfy the condition because a point at a fixed distance from (−9,11)(-9, 11) traces a circle of radius 525\sqrt{2}, and the locus of the moving centre (x−7,2x)(x - 7, 2x) is a straight line — a line generally cuts a circle in two points, hence two answers.

The distance formula is simply the Pythagorean theorem applied to the horizontal and vertical gaps between two points, first cast in coordinate form by René Descartes in the 17th century.

Question details

Q24 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Coordinate Geometry
Topic
Distance formula / equation of circle
Bloom's level
apply
Syllabus topic
Unit 3: Coordinate Geometry — Distance Formula: Applications involving a Circle's Centre and Radius
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How is the distance formula used here?

The point (−9,11)(-9, 11) is on the circle, so its distance from the centre (x−7,2x)(x - 7, 2x) equals the radius 525\sqrt{2}. Writing this distance squared equal to (52)2=50(5\sqrt{2})^2 = 50 gives (x+2)2+(2x−11)2=50(x + 2)^2 + (2x - 11)^2 = 50.

Why are there two values of x?

The condition produces the quadratic x2−8x+15=0x^2 - 8x + 15 = 0, which factorises as (x−3)(x−5)=0(x - 3)(x - 5) = 0. Both roots give valid centres, (−4,6)(-4, 6) and (−2,10)(-2, 10), each exactly 525\sqrt{2} from (−9,11)(-9, 11). The word 'value(s)' signals both are expected.

What is a common mistake in this problem?

The biggest slip is the double negative in (x−7)−(−9)(x - 7) - (-9): it equals x+2x + 2, not x−16x - 16. The second is forgetting that (52)2=50(5\sqrt{2})^2 = 50, not 525\sqrt{2} or 10.