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

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) (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=
(x1,y1)=(-1,3) and (x2,y2)=(3,0)
r=
r=
r=
r=
r=
Now that we have a center (h,k)=(-1,-3) and a radius r=5.
The standard form of a circle: , (h,k) is the center, and r is the radius.

:
Happy Calculating!!!
Answer by venugopalramana(3286) (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
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