What is the standard form of the equation of a circle with a center (-3, -4) and a radius of 3?

1 Answer
Mar 2, 2018

Answer:

# "The equation of the circle with center" \ \ ( -3, -4 ) \qquad "and" \qquad "radius" \ \ 3, "in standard form, is:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad ( x + 3 )^2 + ( y +4 ) ^2 \ = \ 9. #

Explanation:

# "Recall that the formula for the equation of a circle in standard" # # "form is:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad ( x - h )^2 + ( y - k ) ^2 \ = \ r^2; \qquad \qquad \qquad \qquad \qquad \qquad \qquad (I) #

# \qquad \qquad "where:" \qquad \qquad "center" \ = \ ( h, k ), #

# \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ "radius" \ = \ r. #

# "For the circle we are given, we have:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad "center" \ = \ ( \overbrace{ -3 }^h, \overbrace{ -4 }^k ), #

# \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ "radius" \ = \ \overbrace{ 3 }^r. #

# "So, substituting these values into (I), we get:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad ( x - h )^2 + ( y - k ) ^2 \ = \ r^2; #

# \qquad \qquad \qquad \qquad \qquad ( x - ( -3 ) )^2 + ( y - ( -4 ) ) ^2 \ = \ 3^2; #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad ( x + 3 )^2 + ( y +4 ) ^2 \ = \ 9. #

# "This is the equation of our given circle, in standard form." #

# "Summarizing:" #

# "The equation of the circle with center" \ \ ( -3, -4 ) \qquad "and" \qquad "radius" \ \ 3, "in standard form, is:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad ( x + 3 )^2 + ( y +4 ) ^2 \ = \ 9. #