SOLUTION: Find the standard equation of the circle that satisfies the conditions. Center (-1,-3), passing through the point (3,0)

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Question 61514This question is from textbook College Algebra with modeling and visualization
: Find the standard equation of the circle that satisfies the conditions.
Center (-1,-3), passing through the point (3,0) This question is from textbook College Algebra with modeling and visualization

Found 2 solutions by funmath, venugopalramana:
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Find the standard equation of the circle that satisfies the conditions.
Center (-1,-3), passing through the point (3,0)
In order to write an equation for a circle you need a center and a radius. You're missing a radius, but you can find it with the information that they gave you. The radius is the distance from the center to any point of the circle.
The distance formula is r=highlight%28d=sqrt%28%28x2-x1%29%5E2%2B%28y2-y1%29%5E2%29%29
(x1,y1)=(-1,3) and (x2,y2)=(3,0)
r=d=sqrt%28%283-%28-1%29%29%5E2%2B%280-3%29%5E2%29
r=d=sqrt%28%284%29%5E2%2B%28-3%29%5E2%29
r=d=sqrt%2816%2B9%29
r=d=sqrt%2825%29
r=d=5
Now that we have a center (h,k)=(-1,-3) and a radius r=5.
The standard form of a circle: %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2, (h,k) is the center, and r is the radius.
%28x-%28-1%29%29%5E2%2B%28y-%28-3%29%29%5E2=5%5E2
highlight%28%28x%2B1%29%5E2%2B%28y%2B3%29%5E2=25%29
:
Happy Calculating!!!

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Find the standard equation of the circle that satisfies the conditions.
Center (-1,-3), passing through the point (3,0)
EQN. OF CIRCLE WITH CENTRE AT (H,K) AND RADIUS = R IS
(X-H)^2 + (Y-K)^2 = R^2
(X+1)^2 +(Y+3)^2 =R^2
IT PASSES THROUGH (3,0)...SO,
4^2 + 3^2 =R^2 = 25
HENCE EQN. IS
(X+1)^2 + (Y+3)^2 = 25