The vector equation of the line through the origin perpendicular to the plane is
.. (x, y, z) = t(2, 3, -1)
so the point of interest will satisfy
.. 2(2t) +3(3t) -(-t) = -3
.. 14t = -3
.. t = -3/14
and its coordinates are (x, y, z) = (-3/7, -9/14, 3/14)
I think this might be the answer since (a+b) equals 4, and ab= 3. I'm not sure.
a3 + b3
= (4)3+4(3)
= 12+12
= 24
Hello,
x^2+y^2+8x-6y+21=0
Rewritten in the form of a standard circle equation:
(x-(-4))^2+(y-3)^2=2^2
Therefore, the circle properties are:
(a,b)=(-4,3),r=2
The radius of the circle whose equation is <span>x^2 + y^2 + 8x - 6y + 21 =0 is 2.
Faith xoxo</span>
