Pete R. answered • 04/16/14
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1. To find the line tangent to a circle, we need to find the slope of that line.
A line tangent to a circle is perpendicular to the line that connects the point of tangency and the center.
So, in problem 1a, the slope of the line connecting the point of tangency to the center is:
m = 1 - (-3) which equals -1
-3 - 1
Now, the slope perpendicular to this line (which is the slope of the tangent line) is -1/m1
So, mp = -1/-1 which equals 1
Now, we have the slope of the tangent line AND the point of tangency.
Using y-y1 = m(x-x1), we can solve for the equation of the tangent line.
y - (-3) = 1(x - 1) ⇒ y = x - 4
Solve the rest of #1 the same way
NOTE: The center of circle with equation (x-h)2 + (y-k)2 = r2 is (h,k)
2) The standard form equation of a sphere is (x-x0)2 + (y-y0)2 + (z-z0)2 = R2 where (x0,y0,z0) is the center of the sphere.
Since we have the center, the formula so far is: (x-2)2 + (y+5)2 + (z-4)2 = R2
The radius of a sphere is the distance from the center to a point on the sphere. The distance formula in 3-space is: R2 = (x1 - x2)2 + (y1 - y2)2 + (z1 - z2)2
Plugging in, we have: R2 = (2 - 3)2 + (-5 - -3)2 + (4 - 2)2 ⇒ R2 = 9
So, final solution is: (x-2)2 + (y+5)2 + (z-4)2 = 9
3. Standard form is (x-x0)2 + (y-y0)2 + (z-z0)2 = R2.
We just need to get 2x2 + 2y2 + 2z2 - 8x + 20y - 16z - 38 = 0 to look like this
First, notice that there is no coefficient in the standard form, so we need to get rid of the leading coefficient.
Divide both sides by 2 and we have:
x2 + y2 + z2 - 4x + 10y - 8z - 19 = 0 (note: if there were different coefficients, this would not be a sphere)
Rewrite to where variables are together (but we can't add since we can't add x's to x2's)
x2 - 4x + y2 + 10y + z2 = 19 (put the constant on the other side)
We are now going to make each variable expressions (x,y,z) and make it into a perfect square by completing the square.
x2 - 4x + ____ = (x + ____)2
Halve the b value (coefficient of the x term - which is -4) and you get -2 - so we know that we will end up with:
(x - 2)2
If you FOIL this, we would have x2 - 4x + 4. So, we have to add 4 to both sides of the equation (you can't just add 4 to one side for no reason)
x2 - 4x + 4 + y2 + 10y + z2 = 19 + 4 OR
(x2 - 2)2 + y2 + 10y + z2 = 19 + 4
Do the same for the y2 + 10x
Halve 10 and we get 5
(y + 5)2 = y2 + 10y + 25
So add 25 to both sides
x2 - 4x + 4 + y2 + 10y + 25 + z2 = 19+4+25 OR
(x - 4)2 + (y + 5)2 + z2 = 19+4+25
NOW, the z2 doesn't have a z term, so we don't need to complete the square (think of it as (z-0)2)
So, the final equation is:
(x-2)2 + (y+5)2 + z2 = 48
Center of sphere is (2,-5,0) and the radius would be √48
4) Intersection - just plug in the equation of the plane and solve
4a) x2 + (y+5)2 + (z-3)2 = 2500 and y=45
Note: the radius of the sphere, Rsphere, is 50 (√2500). This is necessary for later
Plug in y=45 for the equation of the sphere
x2 + (45+5)2 + (z-3)2= 2500
x2 + 2500 + (z-3)2 = 2500
x2 + (z-3)2 = 0 equation of a circle (x-x0)2 + (z-z0)2=r2 where the center is at (x0,z0) with radius r.
So, center of circle is at (0,45,3) (since y=45 was given) and a rcircle = 0
IF:
A) Rsphere = rcircle, then the intersection would be the GREAT CIRCLE
B) 0<rcircle<Rsphere, then the intersection is a CIRCLE
C) rcircle = 0, then the intersection is a POINT
D) rcircle < 0, then there is NO INTERSECTION
In 4a, since r=0, the intersection is a point at (0,45,3)
Whew!!
