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

"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.