what is the equation of the circle with center (-6,7) that passes through the point (4,-2) ?
Answer choices:
(x+6)^2 + (y-2)^2=9
(x-3)^2 + (y+2)^2=9
(x-3)^2+(y+2)^2=3
(x=3)^2=(y-2)^2=3
clearly, the circle is
(x+6)^2 + (y-7)^2 = r^2
So, I suspect a typo. Check your coordinates again, and recall that a circle with center at (h,k) is
(x-h)^2 + (y-k)^2 = r^2
Once you fix your error, just find how far away it is from (4,-2).
To find the equation of the circle with a given center and a point it passes through, you can use the general equation of a circle:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) represents the center of the circle, and r is the radius.
In this case, the center of the circle is (-6, 7), so h = -6 and k = 7. The circle passes through the point (4, -2), which means it is also on the circumference, so we can substitute these values into the equation.
Substituting the values into the equation, we get:
(4 - (-6))^2 + (-2 - 7)^2 = r^2
(4 + 6)^2 + (-2 - 7)^2 = r^2
10^2 + (-9)^2 = r^2
100 + 81 = r^2
181 = r^2
So the radius squared (r^2) is 181.
Now we can substitute the values of the center and the radius squared into the equation:
(x - (-6))^2 + (y - 7)^2 = 181
(x + 6)^2 + (y - 7)^2 = 181
Therefore, the correct equation of the circle is (x + 6)^2 + (y - 7)^2 = 181. None of the answer choices provided match the correct equation.