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

User Svennergr
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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
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6.0k 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
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