Question

Determine the equation of the circle with center (0, -3) containing the point
(√32,-2).

Answer 1

x² +(y +3)² = 33

Explanation:

You want the equation of the circle with center (0, -3) through point (√32, -2).

Circle equation

The circle with center (h, k) and radius r has equation ...

(x -h)² +(y -k)² = r²

The value of r² can be found by substituting the given point into the left-side expression.

(x -0)² +(y -(-3))² = (√32 -0)² +(-2 -(-3))²

x² +(y +3)² = 32 +1

Th equation of the circle is ...

x² +(y +3)² = 33

Answer 2

• x² + (y + 3)² = 33

======================

The standard form of a circle is as follows:

• (x - h)² + (y - k)² = r², where (h, k) is the center of the circle and r is its radius.

We are given that (h, k) = (0, - 3) and we need to find the value of r².

We know that radius is the distance from the center to a point on the circle.

Recall the distance formula:

• 
  `d=√((x_2-x_1)^2+(y_2-y_1)^2)`

Here, (x₁, y₁) and (x₂, y₂) refer to the coordinates of two points in space.

Substitute the given coordinates and find the radius:

• 
  `r =\sqrt{ (√(32) - 0)^2 + ((-2) - (-3))^2} =√(32+1) =√(33)`

Now, we can write the equation of a circle in standard form:

• 
  `(x - 0)^2 + (y + 3)^2 = (√(33) )^2`

or

• 
  `x ^2 + (y + 3)^2 =33`