Math, asked by DamienVesper, 1 year ago

find the equation of the circle passing through (1,-2) and (4, -3) whose centre lies on the line 3x+4y=7

Answers

Answered by Swarup1998
1

Finding the radius of the circle:

Given that the circle passes through the point (1, - 2) and the centre of the circle lines on the straight line 3x + 4y = 7.

Then the distance of the straight line 3x + 4y = 7 from the point (1, - 2) is given by

| 3 (1) + 4 (- 2) - 7 | / {√(3² + 4²)} units

= | 3 - 8 - 7 | / 5 units

= | - 12 | / 5 units

= 12/5 units, this is the radius of the required circle.

Finding the centre of the circle:

Let the centre of the circle is at (p, q).

Then (p, q) is equidistant from the points (1, - 2) and (4, - 3)

This gives:

√{(p - 1)² + (q + 2)²} = √{(p - 4)² + (q + 3)²}

or, √(p² - 2p + 1 + q² + 4q + 4) = √(p² - 8p + 16 + q² + 6q + 9)

or, p² - 2p + 1 + q² + 4q + 4) = p² - 8p + 16 + q² + 6q + 9

or, - 2p + 4q + 5 = - 8p + 6q + 25

or, 6p - 2q = 20

or, 3p - q = 10 ..... (1)

Again given that the centre (p, q) lies on the straight line 3x + 4y = 7. Then

3p + 4q = 7 ..... (ii)

Subtracting (i) from (ii), we get

3p + 4q - 3p + q = 7 - 10

or, 5q = - 3

or, q = - 3/5

Putting q = - 3/5 in (1), we get

3p - (- 3/5) = 10

or, 3p + 3/5 = 10

or, 3p = 10 - 3/5

or, 3p = (50 - 3)/5

or, p = 47/15

Therefore the centre of the circle is at (47/15, - 3/5).

Finding the equation of the circle:

Therefore the equation of the circle whose radius is 12/5 units and centre is at (47/15, - 3/5), is given by

(x - 47/15)² + (y + 3/5)² = (12/5)²

Read more on Brainly:

  1. find the equation of the circle with centre at (-3,-3) and passing through the point (-3,-6) - https://brainly.in/question/14970215
  2. Find the equation of the circle passing through the point of intersection of the lines X+3y = 0 & 2x-7y=0 and whose cent... - https://brainly.in/question/14459734
Answered by atharvacjoshi
0

Step-by-step explanation:

Let the center of the circle be (h,k)

Since the center is equidistant from all the points on the circle,

(h−1)  

2

+(k+2)  

2

=(h−4)  

2

+(k+3)  

2

 

⇒h  

2

−2h+1+k  

2

+4k+4=h  

2

−8h+16+k  

2

+6k+9

⇒6h−2k−20=0 or 3h−k−10=0    ........(1)

The center also satisfies 3h+4k−7=0

Finding the intersection of these lines, we get

3h−k−10 = 3h+4k−7

Put value in equation (1)

5k+3=0 or k=  

5

−3

​  

   

∴3h=10−  

5

3

​  

=  

5

47

​  

 or h=  

15

47

​  

 

The radius will be  

(  

15

47

​  

−1)  

2

+(  

5

−3

​  

+2)  

2

 

​  

=  

(  

15

32

​  

)  

2

+(  

5

7

​  

)  

2

 

​  

=  

225

1024+441

​  

 

​  

=  

225

1465

​  

 

​  

 

The circle equation is therefore (x−  

15

47

​  

)  

2

+(y+  

5

3

​  

)  

2

=  

45

293

​  

Similar questions