Question

The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is ______.

Answer 1

The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is x² + y² – 46x + 22y = 0

Explanation:

Let AB represents 2y = 3x   ......(i)

BC represents 3y = 4x  ......(ii)

And AC represents y = x + 2  .......(iii)

From equation (i) and (ii)

2y = 3x

⇒ y = `(3x)/2`

Putting the value of y in equation (ii) we get

`3((3x)/2) = 4x`

⇒ 9x = 8x

⇒ x = 0 and y = 0

From equation (i) and (iii) we get

y = x + 2

Putting y = x + 2 in equation (i) we get

2(x + 2) = 3x

⇒ 2x + 4 = 3x

⇒ x = 4 and y = 6

∴ Coordinates of B = (0, 0)

From equation (i) and (iii) we get

y = x + 2

Putting y = x + 2 in equation (i) we get

2(x + 2) = 3x

⇒ 2x + 4 = 3x

⇒ x = 4 and y = 6

∴ Coordinates of A = (4, 6)

Solving equation (ii) and (iii) we get

y = x + 2

Putting the value of y in equation (ii) we get

3(x + 2) = 4x

⇒ 3x + 6 = 4x

⇒ x = 6 and y = 8

∴ Coordinates of C = (6, 8)

It implies that the circle is passing through (0, 0), (4, 6) and (6, 8).

We know that the general equation of the circle is

x² + y² + 2gx + 2fy + c = 0   ......(i)

Since the points (0, 0), (4, 6) and (6, 8) lie on the circle then

0 + 0 + 0 + 0 + c = 0

⇒ c = 0

16 + 36 + 8g + 12f + c = 0

⇒ 8g + 12f + 0 = – 52

⇒ 2g + 3f = – 13   ......(ii)

And 36 + 64 + 12g + 16f + c = 0

⇒ 12g + 16f + 0 = – 100

⇒ 3g + 4f = – 25  .......(iii)

Solving equation (ii) and (iii) we get

     2g + 3f = – 13
     3g + 4f = – 25
⇒  6g + 9f = – 39
     6g + 8f = – 50
     (–)   (–)    (+)     
               f  =    11

Putting the value of f in equation (ii) we get

2g + 3 × 11 = – 13

⇒ 2g + 33 = – 13

⇒ 2g = – 46

⇒ g = – 23

Putting the values of g, f and c in equation (i) we get

x² + y² + 2(– 23)x + 2(11)y + 0 = 0

⇒ x² + y² – 46x + 22y = 0