I need to find the radius of a circle given by, x^2 - 10x + y^2 + 8y = 16

Question

rounded to 1 decimal place

Answer 1

For problems of this type, you need to complete the square in the"x department" and in the "y department..

 

In the x department, half of the -10 (coefficient of x) is -5; you square that (25) so you add 25 to both sides of the equation.

 

In the y department, half of the 8 (coefficient of y) is 4; you square that (16) so you add 16 to both sides of the equation; it now looks like

x² -10y + 25 + y² +8y + 16  =  16 + 25 + 16

and this process GUARANTEES that you have two perfect square trinomials; factoring gives

 

(x - 5)² + (y + 4)² = 57.

 

Comparing this to the general equation of the circle of radius r and center at (h, k): (x-h)² +(y-k)² = r², 

you now see that h = 5, k = -4 and radius is square root of 57.

 

Ask your calculator what the square root of 57 is, and round appropriately, as instructed

Answer 2

Hello Nick,

 

You are given the general form of the equation of a circle

 

X² -10X + Y² -8Y = 16

 

What you need to do is complete the square. to write (X + A)²    and (Y +B)²  . So we need to be able to add something to the left side of the equation to make this possible. To do this take the coefficients of the X and the Y, divide by 2 and square it

 

 

X² -10X             +Y² + 8Y          = 16

 

take -10, divide by 2 you get -5 square it to get 25

 

A = (-10/2) = -5

B = (8/2)    = 4

 

if we add A and B to the left-hand side of our equation of the circle, we must add them to the right-hand side to leave the equation unchanged.

 

X² -10X + (-5)² +Y² + 8Y + (4)² = 16 + (-5)² + (4)²

 

(X - 5)² + (Y + 4)² = 57

 

so the radius is 

 

r² = 57 or r =sqrt(57) =7.5