The point (4,7) lies on the circle centered at (-3,-2), how do you find the equation of the circle in standard form?

1 Answer
Feb 1, 2016

Answer:

# (x + 3 )^2 + (y + 2)^2 = 130 #

Explanation:

the equation of a circle in standard form is :

# (x - a )^2 + (y - b )^2 = r^2 #

where (a , b ) is the centre and r , the radius

In this question the centre is given but require to find r

the distance from the centre to a point on the circle is radius.

calculate r using # color(blue)(" distance formula ") #

which is : # r = sqrt( (x_2 - x_1 )^2 + (y_2 - y_1 )^2) #

using # (x_1 , y_1 ) = (-3,-2))color(black)(" and") (x_2 , y_2) = (4,7)#

then # r =sqrt(4-(-3)^2+(7-(-2)^2)) =sqrt(49+81) =sqrt130#

circle equation using centre =(a , b ) = (-3 , -2) , r #=sqrt130#

# rArr (x+3)^2 +(y+2)^2 = 130 #