The center of a circle is located at (−2, 7) . The radius of the circle is 2. What is the equation of the circle in general form?

Question

5836 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-11T14:45:28

D

D.
It is explained already by another contributor

Answer 2 · 2017-05-11T14:52:29

The answer is the last one, $x^{2} + y^{2} + 4 x - 14 y + 49 = 0$ $x^2+y^2+4x-14y+49=0$

We need

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$

${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2=a^2-2ab+b^2$

The equation of a circle, center $(h, k)$ $(h,k)$ and radius $= r$ $=r$ is

${(x - h)}^{2} + {(y - h)}^{2} = r^{2}$ $(x-h)^2+(y-h)^2=r^2$

Here, the center is $(- 2, 7)$ $(-2,7)$ and the radius is $= 2$ $=2$

The equation of the circle is

${(x + 2)}^{2} + {(y - 7)}^{2} = 4$ $(x+2)^2+(y-7)^2=4$

$x^{2} + 4 x + 4 + y^{2} - 14 y + 49 - 4 = 0$ $x^2+4x+4+y^2-14y+49-4=0$

$x^{2} + y^{2} + 4 x - 14 y + 49 = 0$ $x^2+y^2+4x-14y+49=0$