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 in the standard form of a circle ( x − h ) 2 + ( y − k ) 2 = r 2 .
Complete the square for both x and y terms to get the equation in standard form.
Identify the center ( h , k ) from the standard equation.
The center of the circle is ( 6 , 1 ) .
Explanation
Analyze the problem and rewrite the equation. 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 ) is the center of the circle and r is the radius.
Complete the square for x terms. First, let's group the x terms and y terms together: ( x 2 − 12 x ) + ( y 2 − 2 y ) + 12 = 0 .
Now, 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 can rewrite x 2 − 12 x as ( x − 6 ) 2 − 36 .
Complete the square for y terms. Next, 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 can rewrite y 2 − 2 y as ( y − 1 ) 2 − 1 .
Substitute back into the original equation. Now, substitute these back into the original equation: ( x − 6 ) 2 − 36 + ( y − 1 ) 2 − 1 + 12 = 0
Simplify the equation. Simplify the equation: ( x − 6 ) 2 + ( y − 1 ) 2 = 36 + 1 − 12
( 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 . 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 fields. For instance, in GPS technology, your location is determined by finding the intersection of circles from multiple satellites. The center and radius of these circles, derived from the signal travel time, pinpoint your exact coordinates on Earth. This principle extends to fields like astronomy, where the orbits of celestial bodies are often described using conic sections, including circles.
