Question

Determine the equation of the circle with center (0,−6) containing the point (−\sqrt{28 },3)

Answer 1

The equation of the circle with center (0, -6) containing the point (-√28, 3) is x^2 + (y + 6)^2 = 109.

Explanation:

The equation of a circle with center (h, k) and radius r is given by the formula:

(x - h)^2 + (y - k)^2 = r^2

In this case, the center of the circle is (0, -6), which means that h = 0 and k = -6. We also know that the circle contains the point (-√28, 3), which means that this point is on the circle and satisfies the equation above.

To find the radius r, we can use the distance formula between the center of the circle and the given point:

r = sqrt((0 - (-√28))^2 + (-6 - 3)^2) = sqrt(28 + 81) = sqrt(109)

Substituting h, k, and r into the equation of the circle, we get:

x^2 + (y + 6)^2 = 109

Therefore, the equation of the circle is x^2 + (y + 6)^2 = 109.

Answer 2

I did not get Your Question Probably But the way I see it the answer is:

The equation of a circle with center (h, k) and radius r can be written as:

(x - h)^2 + (y - k)^2 = r^2

In this case, the center of the circle is (0, -6). To find the radius, we can use the distance formula between the center and the given point (-√28, 3):

r = √((x₂ - x₁)^2 + (y₂ - y₁)^2)

Plugging in the values:

r = √((-√28 - 0)^2 + (3 - (-6))^2)

Simplifying:

r = √(28 + 81)

r = √109

Therefore, the equation of the circle with center (0, -6) and containing the point (-√28, 3) is:

(x - 0)^2 + (y + 6)^2 = (√109)^2

Simplifying further:

x^2 + (y + 6)^2 = 109