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