In the plane, we have everywhere. So in the equation of the sphere, we have
which is a circle centered at (2, -10, 0) of radius 4.
In the plane, we have , which gives
But any squared real quantity is positive, so there is no intersection between the sphere and this plane.
In the plane, , so
which is a circle centered at (0, -10, 3) of radius .
The correct answer should be letter (B)
Answer:
D
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Here (h, k) = ( - , ) and r = , thus
(x - (- ))² + (y - )² = ( )², that is
(x + )² + (y - )² = → D
This is called difference of squares. It's like
a^2-b^2=(a-b)(a+b)
t^2-25=(t-5)(t+5)
Hope this helps :)