What is the center of a circle whose equation is $x^2+y^2-12 x-2 y+12=0$?
A. $(-12,-2)$
B. $(-6,-1)$
C. $(6,1)$
D. $(12,2)$
Answer (1)
Rewrite the given equation x 2 + y 2 − 12 x − 2 y + 12 = 0 by grouping x and y terms: ( x 2 − 12 x ) + ( y 2 − 2 y ) + 12 = 0 .
Complete the square for x terms: ( x − 6 ) 2 − 36 .
Complete the square for y terms: ( y − 1 ) 2 − 1 .
Substitute back and simplify to standard form: ( x − 6 ) 2 + ( y − 1 ) 2 = 25 , thus the center is ( 6 , 1 ) .
Explanation
Analyze the problem and rewrite in standard form We are given the equation of a circle: x 2 + y 2 − 12 x − 2 y + 12 = 0 . Our goal is to find the center of this circle. To do this, we will rewrite the equation in the standard form of a circle's equation, which is ( x − h ) 2 + ( y − k ) 2 = r 2 , where ( h , k ) represents the center of the circle and r is the radius.
Complete the square for x terms First, we group the x and y terms together: ( x 2 − 12 x ) + ( y 2 − 2 y ) + 12 = 0 .
Next, we complete the square for the x terms. To complete the square for x 2 − 12 x , we take half of the coefficient of the x term, which is − 12/2 = − 6 , and square it: ( − 6 ) 2 = 36 . So, we add and subtract 36 within the equation: ( x 2 − 12 x + 36 − 36 ) + ( y 2 − 2 y ) + 12 = 0 ( x − 6 ) 2 − 36 + ( y 2 − 2 y ) + 12 = 0 .
Complete the square for y terms Now, we complete the square for the y terms. To complete the square for y 2 − 2 y , we take half of the coefficient of the y term, which is − 2/2 = − 1 , and square it: ( − 1 ) 2 = 1 . So, we add and subtract 1 within the equation: ( x − 6 ) 2 − 36 + ( y 2 − 2 y + 1 − 1 ) + 12 = 0 ( x − 6 ) 2 − 36 + ( y − 1 ) 2 − 1 + 12 = 0 .
Simplify the equation Now, we simplify the equation by combining the constants: ( x − 6 ) 2 + ( y − 1 ) 2 − 36 − 1 + 12 = 0 ( x − 6 ) 2 + ( y − 1 ) 2 − 25 = 0 ( x − 6 ) 2 + ( y − 1 ) 2 = 25 .
Identify the center of the circle The equation is now in the standard form ( x − 6 ) 2 + ( y − 1 ) 2 = 25 , which can be written as ( x − 6 ) 2 + ( y − 1 ) 2 = 5 2 . Comparing this with the standard form ( x − h ) 2 + ( y − k ) 2 = r 2 , we can identify the center of the circle as ( h , k ) = ( 6 , 1 ) .
State the final answer Therefore, the center of the circle is ( 6 , 1 ) .
Examples
Understanding the equation of a circle is crucial in various real-world applications. For instance, when designing a circular garden or a roundabout, knowing the center and radius helps in accurately planning the layout and dimensions. Similarly, in fields like astronomy, the orbits of celestial bodies can often be approximated as circles or ellipses, and determining the center and other parameters is essential for predicting their paths.
