Q.385. Polynomial equation with coefficients in A.P.

Question 385.
Find a necessary and sufficient condition on a, b, c
such that the roots of x³ + ax² + bx + c = 0 are in arithmetic progression.

Answer 385.
Let the roots be m-d, m and m+d
=> m-d + m + m+d = - a
=> m = - a/3
m(m-d) + m(m+d) + (m-d)(m+d) = b
=> 3m^2 - d^2 = b
=> d^2 = a^2/3 - b
=> roots are
-a/3 - √(a^2/3-b), - a/3 and -a/3 + √(a^2/3-b)
=> product of the roots
(-a/3 - √(a^2/3 - b) * (-a/3) * (-a/3 + √(a^2/3 - b) = - c
=> - (a/3) * (a^2/9 - a^2/3 + b) = - c
=> (a/3) (b - 2a^2/9) = c
=> c = (a/27) (9b - 2a^2)
This is the necessary and sufficient condition for the roots to be in A.P.
Sufficient because the roots of the equation with the above value of c are the ones found as above for which refer to the following Wolfram Alpha link:
http://www.wolframalpha.com/input/?i=x%C2%B3+%2B+ax%C2%B2+%2B+bx+%2B+%28a%2F27%29%289b-2a%5E2%29+%3D+0

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Q.384. Center and radius of the circle in 3-D.

Question 384.
Find the coordinates of center and radius of the circle of intersection of 
the sphere x²+y²+z²=9 with the plane 3x+4y+5z=5,

Answer 384.
The equation of the family of spheres through the given sphere and the given plane is
x^2 + y^2 + z^2 - 9 + k (3x + 4y + 5k - 5) = 0.

The centre of the intersecting circle is the centre of the sphere from the above family which lies on the plane.
The centre of the sphere = (-3k/2, - 4k/2, - 5k/2)

Plugging in the eqn. of the plane
=> - 9k/2 - 16k/s - 25k/2 - 5 = 0
=> k = -1/5
=> eqn. of the sphere containing the intersecting circle as its large circle is
x^2 + y^2 + z^2 - 9 - (1/5) (3x + 4y + 5z - 5) = 0
Its centre is (3/10, 2/5, 1/2)
Its radius
= √[(3/10)^2 + (4/10)^2 + (5/10)^2 + 9 - 1]
= √[1/2 + 8]
= √(17/2)

Alternate method to find the radius is
radius of the sphere
r = 3
perpendicular distance from the centre (0, 0, 0) of the sphere to the plane
p = 5/√(3^2 + 4^2 + 5^2) = 1/√2
=> radius of the circle of intersection
= √(r^2 - p^2)
= √(9 - 1/2)
= √(17/2).

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