15 POINTS 1.) name two planes that intersect in WX in the figure to the right

If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle

stira [4]

Answer:

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

 or 

The equation of the circle x² + 2 x + y² - 4 y = 15

Step-by-step explanation:

Given points end Points are p(-3,-2) and q( 1,6)

The distance of two points formula

P Q = $\sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2} }$

P Q = $\sqrt{1 - (-3)^{2} + ((6 -(-2))^{2} }$

P Q = $\sqrt{16+64} = \sqrt{80}$

The diameter 'd' = 2 r

2 r = √80

= $\sqrt{16 X 5}$

= $4 \sqrt{5}$

r = 2√5

Mid-point of two end points

$(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )$

= (-1 ,2)

Mid-point of two end points = center of the circle

(h,k) = (-1 , 2)

The equation of the circle

(x -h )² +(y-k)² = r²

(x -(-1) )² +(y-(2))² = (2(√5))²

x² + 2 x + 1 + y² - 4 y + 4 = 20

x² + 2 x + y² - 4 y = 20 -5

x² + 2 x + y² - 4 y = 15

Final answer:-

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

 or 

The equation of the circle x² + 2 x + y² - 4 y = 15

6 0

3 years ago

Find the arc-length parametrization of the curve that is the intersection of the elliptic cylinder x 2 + y 2/2 = 1 and the plane

storchak [24]

Answer:

$f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)$

0 ≤ Ф ≤ 4π.

Step-by-step explanation:

since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))

the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.

We take

$f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)$

Note that