Answer to Question #320759 in Geometry for Alina

Question

Find the equation (formula) of a circle with radius r and center C(h,k) and if the Center of a circle is at (3,-1) and a point on the circle is (-2,1) find the formula of the circle.

 

Find the equation (formula) of a sphere with radius r and center C(h, k, l) and show that x²Â + y²Â + z²Â - 6x + 2y + 8z - 4 = 0 is an equation of a sphere. Also, find its center and radius.

 

 

Answer 1

The equation of the circle:

$(x-h)^2+(y-k)^2=r^2$

$C(h,k)=(3,-1)$

Point: $(-2;1)$ .

$(-2-3)^2+(1-(-1))^2=r^2$

$25+4=r^2$

$r=\sqrt {29}$

the equation of the circle:

$(x-3)^2+(y+1)^2=29$ .

The equation of the sphere:

$(x-h)^2+(y-k)^2+(z-l)^2=r^2$

$x^2 + y^2 + z^2 - 6x + 2y + 8z - 4 = 0$

$(x^2 -6x+9)-9+ (y^2 +2y+1)-1+$$( z^2+8z+16)-16 -4=0$

$(x-3)^2+(y+1)^2+(z+4)^2-30=0$

The equation of the sphere

$(x-3)^2+(y+1)^2+(z+4)^2=30$ .

The centre of the sphere:

$C(h,k,l)=(3,-1,-4)$ .

Radius:

$r=\sqrt {30}$ .