find the center and radius of the circle whose equation is x^2-24x+49=-y^2+14y

Question

find the center and radius of the circle whose equation is x^2-24x+49=-y^2+14y

Answer 1

Hi Riley,

 

A circle of radius r centered at (x,y) = (h,k) is given by

 

(x - h)² + (y - k)² = r²

 

You have x² - 24x + 49 = -y² + 14y

 

Then, through rearranging, we have

 

x² - 24x + y² -14y + 49 = 0

 

Let's complete the square for x and for y.

 

Remember, a perfect square looks like (x + a)² = x² + 2ax + a²

 

So for x² - 24x to be part of a perfect square, So, -24 = 2a, so a = -12. Then a² = 144.

 

So, x² - 24x + 144 = (x - 12)².

 

For the y's, 2a = -14, so a = -7. Then, a² = 49. So, a perfect square for y would be y2 - 14y + 49 = (y - 7)².

 

So, to make this work, we need to add and subtract 144 and 49. So, we get

 

x² - 24x +144 - 144 + y² -14y + 49 - 49 + 49 = 0

 

Then, we have

 

(x - 12)² -144 + (y - 7)² - 49 + 49 = 0

 

The 49's cancel. Then, we can add 144 to each side of the equation to get

 

 (x - 12)² + (y - 7)² = 144

 

Or

 

(x - 12)² + (y - 7)² = 12²

 

This equation describes a circle of radius 12 centered about (h, k) = (12, 7).

Answer 2

 

x^2 - 24x + 49 = -y^2 + 14y

 

x^2 - 24x + y^2 - 14y = -49  <--- moves all the vars to the left and the constant to the right

 

(X^2 - 24x + 144) + (y^2 - 14y + 49) = -49 + 49 + 144 <--- completes the square

 

(x-12)^2 + (y-7)^2 = 144 <--- this is the equation of the circle

 

the center is (12,7)

the radius is 12

 

As a check, the point (12,-5) is on the edge of the circle.

So are (12,19)  ,    (0,7) and (24,7)

All of these points satisfy the equation of the circle in bold above.

 

Therefore, the answers given in bold above are proven to be correct.

The center is (12,7) and the radius is 12.