How to Find Radius and Center by Circle General Equation (METHOD 2)

Click here for METHOD 1

EXAMPLE: Find center and radius of the circle with equation

x² + y² – 8x + 2y + 8 = 0

SOLUTION:

METHOD 2:

Convert the general equation of given circle to standard equation of that circle. Then read the center and radius from that equation.

(Here I am using the technique of completing the square.)

x² + y² – 8x + 2y + 8 = 0       (given)

Step 1: If the squared terms have any co-efficient other than 1 then divide the whole equation by that co-efficient.

Here in the given equation, the co-efficient of x² and y² is 1, so we will skip this step.

Step 2: Group the x-terms, group the y-terms, and take the constant to the right side

x² + y² – 8x + 2y + 8 = 0             (given)

(x² – 8x) + (y² + 2y) = – 8  ———— (1)  

Step 3: Divide the co-efficient of the first-degree term by 2 and square it.

In (x² – 8x),

8x is first-degree term. 8 is co-efficient of x. 

8 ÷ 2 = 4

4² = 16.

In (y² + 2y),

2y is first-degree term. 2 is co-efficient of y.

2 ÷ 2 = 1

1² = 1.

Step 4: Add these constants within parentheses respectively, to complete the square, and add the same constants to the right side to maintain equality.

(x² – 8x) + (y² + 2y) = – 8  ———— (1)

(x² – 8x + 16) + (y² + 2y + 1) = – 8 + 16 + 1 

(x² – 8x + 16) + (y² + 2y + 1) = 9 ———- (2)

Step 5:

As we know the formulae,

a² – 2ab + b² = (a – b)²

a² + 2ab + b² = (a + b)²

Apply these formulae

x² – 8x + 16 = x² – 2(4)x + 4² 

= (x – 4)² 

y² + 2y + 1 = y² – 2y + 1² 

= (y + 1)²

So, equation (2) reduces to

(x – 4)² + (y + 1)² = 9 

(x – 4)² + {y – (-1)}² = 9 —————-(3)

This is the Standard Equation of given circle.

Step 6:

We know,

standard equation of a circle is:

(x – h)² + (y – k)² = r² ————- (4)

where (h, k) is center of a circle and r is the radius.

Compare equation (3) and equation (4), 

we get

h = 4

k = -1

r² = 9 implies r = 3

Hence, a circle with equation

x² + y² – 8x + 2y + 8 = 0

has center at (4, -1) and radius 3.

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