The Coordinates of the Centre of a Circle Are (x − 7, 2x): Find the Value of x if the Circle Passes Through (−9, 11) and Has Radius 5√2 | Step-by-Step Solution

 The coordinates of the centre of a circle are (x − 7, 2x). Find the value(s) of ‘x’, if the circle passes through the point (−9, 11) and has radius 5√2 units.

For a circle, 

distance between centre and any point on circle = radius

using square of distance formula:

 (x₂ − x₁)² + (y₂ − y₁)² = r²

Given 

Centre = (x − 7, 2x)

Point on circle = (−9, 11)

Radius r = 5√2 

 r² = (5√2)² = 25 × 2 = 50

(-9 - (x - 7))² + (11 - 2x)² = 50

(-9 - x + 7)² + (11 - 2x)² = 50

(-x - 2)² + (11 - 2x)² = 50

(x + 2)² + (11 - 2x)² = 50

Expand using identities

(x² + 4x + 4) + (121 - 44x + 4x²) = 50

5x² - 40x + 125 = 50

5x² - 40x + 125 - 50 =0

5x² - 40x + 75 = 0

Divide by 5

x² - 8x + 15 = 0

Factorise:

x² - 5x - 3x + 15 = 0

x(x - 5) - 3(x - 5) = 0

(x - 5)(x - 3) = 0

x = 5 or x = 3

for  more explanation watch the video  

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cbse 10th maths coordinate geometry distance formula previous year question paper 2025 2026

Find a relation between x and y such that the point P(x, y) is equidistant from the points A(5, 3) and B(1, 7).

Find a relation between x and y such that the point P(x, y) is equidistant from the

 points A(5, 3) and B(1, 7)

Given

P(x, y) is equidistant from A and B, 

 PA = PB

squaring both sides

PA² = PB²

Using distance formula to find PA and PB

(x - 5)² + (y - 3)² = (x - 1)² + (y - 7

Expand both sides using identity

x² - 10x + 25 + y² - 6y + 9 = x² - 2x + 1 + y² - 14y + 49

-10x - 6y + 34 = -2x - 14y + 50

-10x + 2x - 6y + 14y = 50 - 34

-8x + 8y = 16

or

-x + y = 2

or 

y = x + 2

watch this video for more details

cbse 10th maths 2025 2026 old board exam question paper coordinate geometry distance formula

Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

 exercise 7.1 co ordinate geometry chapter 7 cbse ncert 10th mathematics distance formula, collinear points using distance formula

3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

let

A= (1, 5)

B= (2, 3)

C= (– 2, – 11) 

 

A= (1, 5)=(x1,y1)

B= (2, 3)=(x2,y2)

 

using distance formula

AB= sqrt{(2-1)^2 +(3-5)^2} =sqrt(5)

B= (2,3)=(x1,y1)

C= (-2, -11)=(x2,y2)

 

using distance formula

BC= sqrt{(-2-2)^2 +(-11-3)^2} =sqrt(212)=2sqrt(53)

 

 

A= (1, 5)=(x1,y1)

C= (-2, -11)=(x2,y2)

 

AC=sqrt{(-2-1)^2 +(-11-5)^2}  =sqrt{265}

clearly AC is the biggest

but AB+BC is clearly not equal to AC

so the points are not collinear

 

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order  

solution

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

solution

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

exercise 7.1

Find the distance between the following pairs of points :(2, 3), (4, 1)

 solution

(ii) (– 5, 7), (– 1, 3)

solution

 

(iii) (a, b), (– a, – b)

solution

2. Find the distance between the points (0, 0) and (36, 15).

 solution 

3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

solution

 

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

Find the distance between the following pairs of points (ii) (– 5, 7), (– 1, 3)

 

 exercise 7.1 co ordinate geometry chapter 7 cbse ncert 10th mathematics distance formula, collinear points using distance formula

Find the distance between the following pairs of points :((ii) (– 5, 7), (– 1, 3)

(-5,7) =(x1,y1)

(-1,3))=(x2,y2)

distance =sqrt{(-1-(-5) )^2 +(3-7)^2} =sqrt(32)=4sqrt(2) units

 

(iii) (a, b), (– a, – b)

 

(a,b) =(x1,y1)

(-a,-b))=(x2,y2)

distance =sqrt{(-a-a)^2 +(-b-b)^2}  =sqrt{4 ( a^2  + b^2 )}

=2sqrt{ ( a^2  + b^2 )}

 

2. Find the distance between the points (0, 0) and (36, 15).

(0,0) =(x1,y1)

(36,15))=(x2,y2)

 

distance ={(36-0)^2 +(15-0)^2}  = sqrt(1521)=39 units

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order  

solution

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

solution

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

exercise 7.1

Find the distance between the following pairs of points :(2, 3), (4, 1)

 solution

(ii) (– 5, 7), (– 1, 3)

solution

 

(iii) (a, b), (– a, – b)

solution

2. Find the distance between the points (0, 0) and (36, 15).

 solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order

 exercise 7.2 co ordinate geometry chapter 7 cbse ncert 10th mathematics section formula , midpoint formula  

 

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order

take 

A= (3, 0)

B=(4, 5)

C= (– 1, 4) 

D=(– 2, – 1)

 

diagonals are AC and BD

 

A= (3, 0) =(x1,y1)

C= (– 1, 4)=(x2,y2)

 

using distance formula

AC= sqrt{(-1-3)^2  + (4-0)^2} =sqrt(32)=4sqrt(2)

B= (4,5) =(x1,y1)

D= (– 2, -1)=(x2,y2)

 

using distance formula

BD=sqrt{ (-2-4)^2  +(-1-5)^2 }  =sqrt(72)=6sqrt(2)

area of a rhombus =(1/2)*AC*BD=24 sq.units

 

 exercise 7.1 co ordinate geometry chapter 7 cbse ncert 10th mathematics distance formula, collinear points using distance formula

Find the distance between the following pairs of points :(2, 3), (4, 1)

(2,3) =(x1,y1)

(4,1))=(x2,y2)

 

using distance formula

 

distance = sqrt{(4-2)^2 +(1-3)^2} =sqrt(8) =2sqrt(2) units.

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

solution

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

(ii) (8, 1), (k, – 4), (2, –5)

solution 

 

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

solution 

 

exercise 7.2

10.Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order  

solution

9. Find the coordinates of the points which divide the line segment joining

 A(– 2, 2) and B(2, 8) into four equal parts. 

solution

 

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP =(3/7) AB and P lies on the line segment AB.

solution

 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1,4)

 solution

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

solution 

 4.Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

 solution

2. Find the coordinates of the points of trisection of the line segment joining

 (4, –1) and (-2,-3)

 

solution

Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the
ratio 2 : 3 

solution

 

exercise 7.1

Find the distance between the following pairs of points :(2, 3), (4, 1)

 

solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.