What is the length of the radius and the coordinates of the center of the circle defined by the equation #(x+7)^2+(y-3)^2=121#?

1 Answer
Mar 14, 2017

Answer:

The radius is #11 (14-3)# and the coordinates of the centre is (#7,3#)

Explanation:

Opening up the equation,
#(x+7)^2 + (y-3)^2 = 121#
#x^2+14x+49+y^2-6y+9 = 121#
#y^2-6y = 63-x^2+14x#

Find the x-intercepts, and the midpoint to find x-line of symmetry,
When #y = 0#,
#x^2-14x-63 = 0#
#x=17.58300524 or x=-3.58300524#

#(17.58300524-3.58300524)/2 = 7#

Find the highest and lowest point and midpoint,
When #x =7#,
#y^2-6y-112=0#
#y = 14 or y = -8#

#(14-8)/2 = 3#

Hence, the radius is #11 (14-3)# and the coordinates of the centre is (#7,3#)