How to find Center and Radius of a Circle from its General Equation (METHOD 2)
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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