Question 242 marksVery Short AnswerSection B

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 and Circle

Official answer

x=3x = 3 or x=5x = 5, obtained by equating the distance from centre (x−7,2x)(x-7, 2x) to the point (−9,11)(-9, 11) with the radius 525\sqrt{2}, giving x2−8x+15=0x^2 - 8x + 15 = 0.

Answer keywords:distance formularadiuscentre of circlequadratic equationcoordinate geometryfactorisation

Marking scheme

  1. 1.1 mark: correctly setting up the distance equation (x+2)2+(2x−11)2=52\sqrt{(x+2)^2 + (2x-11)^2} = 5\sqrt{2} and simplifying to the quadratic x2−8x+15=0x^2 - 8x + 15 = 0.
  2. 2.1 mark: correctly factorising/solving to get both values x=3x = 3 and x=5x = 5.

Hint

Use the distance formula between the centre (x−7,2x)(x-7, 2x) and the given point (−9,11)(-9, 11), and equate it to the radius 525\sqrt{2}.

Quick oral answer

Setting the distance from centre (x−7,2x)(x-7, 2x) to the point (−9,11)(-9, 11) equal to 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, factorising to (x−3)(x−5)=0(x-3)(x-5) = 0, so x=3x = 3 or x=5x = 5.

Common mistakes

  1. 1.Sign error in computing x−7−(−9)x - 7 - (-9), incorrectly simplifying it to x−16x - 16 instead of the correct x+2x + 2.
  2. 2.Squaring 5√2 incorrectly as 25√2 or 10 instead of the correct 50.
  3. 3.Stopping after finding one root of the factorised quadratic and forgetting that both x=3x = 3 and x=5x = 5 are valid solutions.

Analysis and explanation

Combines the Distance Formula with the definition of a circle (all points equidistant from the centre).


Concept

  • Distance from centre (x−7,2x)(x-7, 2x) to a point (−9,11)(-9, 11) on the circle equals the radius 525\sqrt{2}.

Key points

  • Setting up (x+2)2+(2x−11)2\sqrt{(x+2)^2 + (2x-11)^2} = 525\sqrt{2} and squaring leads to the quadratic x2−8x+15=0x^2 - 8x + 15 = 0.
  • Factorising gives x=3x = 3 or x=5x = 5 — both valid since no constraint rules either out.

Common mistakes

  • Sign error simplifying (x−7−(−9))(x-7-(-9)) to (x+2).
  • Arithmetic slip expanding (2x-11)².

Interesting facts

The distance formula (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} used here is a direct application of the Pythagoras theorem to coordinate points, one of the most tested formulas across all Class 10 coordinate geometry problems.

A quadratic equation in coordinate geometry problems like this typically yields two valid answers precisely because a circle's centre condition is symmetric — reflecting the geometric fact that many points can be equidistant (525\sqrt{2} units) from (−9,11)(-9, 11).

525\sqrt{2} as a radius often arises from a right-triangle-like distance calculation (e.g., legs 5 and 5), since 525\sqrt{2} = 50=25+25\sqrt{50} = \sqrt{25+25}.

Question details

Q24 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Coordinate Geometry
Topic
Distance Formula and Circle
Bloom's level
apply
Syllabus topic
Unit 3: Coordinate Geometry — Distance Formula: distance between two points, applied to a circle's centre and a point on it
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why does this question give two values of x?

Because the distance equation, after squaring, becomes a quadratic equation in x, and both roots (x=3x=3 and x=5x=5) satisfy the original condition without any geometric restriction ruling one out.

What formula connects the centre, a point on the circle, and the radius?

The distance formula: if (h,k)(h,k) is the centre and (x1,y1)(x_1,y_1) lies on the circle with radius r, then (x1−h)2+(y1−k)2=r\sqrt{(x_1-h)^2 + (y_1-k)^2} = r.