Fill in the Blanks

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

Solving the given equations,

y = x + 2 & 3y = 4x

3 (x + 2) = 4x

3x + 6 = 4x

x = 6, y = x + 2 = 6 + 2 = 8

Points of intersection is (6, 8)

y = x + 2 & 2y = 3x

2 (x + 2) = 3x

2x + 4 = 3x

x = 4, y = x + 2 = 4 + 2 = 6

Points of intersection is (4, 6)

3y = 4x & 2y = 3x

x = 0 y = 0

Points of intersection is (0, 0)

Expanded form of a circle,

x² + y² + 2gx + 2fy + C = 0, whereas Centre = (-g, -f)

(6, 8)

6² + 8² + 2g(6) + 2f(8) + C = 0

36 + 64 + 12g + 16f + C = 0

16f + 12g + C + 100 = 0

(4, 6)

4² + 6² + 2g(4) + 2f(6) + C = 0

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

12f + 8g + C + 52 = 0

(0, 0)

0² + 0² + 2g(0) + 2f(0) + C = 0

0 + 0 + 0 + 0 + C = 0

C + 0 = 0

C = 0

16f + 12g + 100 = 0

12f + 8g + 52 = 0

Solving the above mentioned equations, simultaneously,

48f + 36g + 300 = 0

48f + 32g + 208 = 0

4g + 92 = 0

g + 23 = 0

g = -23

16f + 12 (-23) + 100 = 0

16f – 276 + 100 = 0

16f = 176

f = 11 & g = -23

Expanded form of a circle,

x² + y² + 2gx + 2fy + C = 0, whereas Centre = (-g, -f)

x² + y² + 2(-23)x + 2(11)y + C = 0

x² + y² - 46x + 22y + C = 0

Hence, the required equation is x² + y² - 46x + 22y + C = 0.