139k views
0 votes
write the standard equation of a circle with the center at (-1,-4) that passes through the point (-1,-1)​

User Svennergr
by
7.5k points

2 Answers

4 votes

Answer:

(x + 1)² + (y + 4)² = 9

Explanation:

The equation of a circle is:

(x - h)² + (y - k)² = r²

where (h, k) is the center and r is the radius.

The center is (h, k) = (-1, -4). The radius is the distance from the center to any point on the circle. (-1, -1) is 3 units above the center, so r = 3.

(x + 1)² + (y + 4)² = 9

User James Martineau
by
8.7k points
1 vote

Answer:

(x+1)^2+(y+4)^2=9

Explanation:

First equation in what I like to call center-radius form is (x-h)^2+(y-k)^2=r^2 where (h,k) is center and r is radius.

So we are given (h,k)=(-1,-4) so we have so far that

(x+1)^2+(y+4)^2=r^2

Passes through the point (-1,-1) means at some point (x,y) takes on the point (-1,-1) so guess what we are going to find r^2 by plugging in that point where x and y are.

(-1+1)^2+(-1+4)^2=r^2

(0)^2+(3)^2=r^2

9=r^2

so the equation of the circle is (x+1)^2+(y+4)^2=9

User Papalagi
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories