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Step2247 [10]
3 years ago
9

PLZ HELP ASAP FEATURES OF A CIRCLE FROM ITS EXPANDED EQUATION LAST ONE

Mathematics
1 answer:
pantera1 [17]3 years ago
5 0
We have the following equation:
 x2 + y2 + 6y - 72 = 0
 Rewriting we have:
 x2 + y2 + 6y = 72
 Complete squares we have:
 x2 + y2 + 6y + (6/2) ^ 2 = 72 + (6/2) ^ 2
 x2 + y2 + 6y + (3) ^ 2 = 72 + (3) ^ 2
 x2 + y2 + 6y + 9 = 72 + 9
 x2 + (y + 3) 2 = 81
 So, the center is:
 (0, -3)
 The radio is:
 r = 9
 Answer:
 
(0, -3)
 
r = 9
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) find a vector parallel to the line of intersection of the planes 5x − y − 6z = 0 and x + y + z = 1.
snow_tiger [21]
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.

Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>

We calculate the cross product as a determinant of (i,j,k) and the normal products

    i   j   k
   5 -1 -6
   1  1  1

=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>

Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0

Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
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