What are the center and radius of the circle defined by the equation $x^2+y^2-6 x+10 y+25=0$?
A. Center $(-3,5)$; radius 3
B. Center $(-3,5)$; radius 9
C. Center $(3,-5)$; radius 9
D. Center $(3,-5)$; radius 3
Answer (1)
Complete the square for both x and y terms in the given equation.
Rewrite the equation in the standard form: ( x − 3 ) 2 + ( y + 5 ) 2 = 9 .
Identify the center as ( h , k ) = ( 3 , − 5 ) .
Determine the radius as r = 9 = 3 .
The center and radius of the circle are ( 3 , − 5 ) ; 3 .
Explanation
Analyze the problem We are given the equation of a circle: x 2 + y 2 − 6 x + 10 y + 25 = 0 . Our goal is to find the center and radius of this circle. To do this, we will rewrite the equation in the standard form of a circle, which is ( x − h ) 2 + ( y − k ) 2 = r 2 , where ( h , k ) is the center and r is the radius.
Complete the square for x terms First, we complete the square for the x terms. We have x 2 − 6 x . To complete the square, we take half of the coefficient of the x term, which is − 6/2 = − 3 , and square it: ( − 3 ) 2 = 9 . So, we add and subtract 9.
Complete the square for y terms Next, we complete the square for the y terms. We have y 2 + 10 y . To complete the square, we take half of the coefficient of the y term, which is 10/2 = 5 , and square it: ( 5 ) 2 = 25 . So, we add and subtract 25.
Rewrite the equation Now we rewrite the equation, adding and subtracting the values we found to complete the square: ( x 2 − 6 x + 9 ) + ( y 2 + 10 y + 25 ) − 9 − 25 + 25 = 0
Simplify the equation We simplify the equation: ( x − 3 ) 2 + ( y + 5 ) 2 − 9 = 0
Isolate the squared terms We move the constant term to the right side of the equation: ( x − 3 ) 2 + ( y + 5 ) 2 = 9
Identify the center and radius Now we can identify the center and radius from the standard form. The center is ( 3 , − 5 ) and the radius is 9 = 3 .
State the final answer Therefore, the center of the circle is ( 3 , − 5 ) and the radius is 3 .
Examples
Understanding the equation of a circle is crucial in various real-world applications. For instance, when designing a circular garden, you need to determine the center and radius to plan the layout accurately. Similarly, in GPS navigation, the location of a device can be determined by finding the intersection of circles with known centers and radii. This concept is also fundamental in fields like astronomy, where the orbits of planets are often approximated as circles or ellipses.
