Question

What is the equation of a circle with center (-3,-1) that contains the point (1,2)?

Answer 1

(x + 3)^2 + (y + 1)^2 = 25

Explanation:

Equation of a circle with center (h, k) and radius, r.

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

The center is (-3, -1), so h = 1, and k = 2.

(x - (-3))^2 + (y - (-1))^2 = r^2

(x + 3)^2 + (y + 1)^2 = r^2

Now we substitute x and y with the values of x and y from the given point, and we solve for r^2.

(1 + 3)^2 + (2 + 1)^2 = r^2

4^2 + 3^2 = r^2

16 + 9 = r^2

r^2 = 25

Now that we know r^2, we substitute it into the equation above.

(x + 3)^2 + (y + 1)^2 = 25

Answer 2

The correct answer is,

(x + 3)² + (y +1)² = 25

Explanation:

It is given that, What is the equation of a circle with center (-3,-1) that contains the point (1,2)

Formula;-

Equation of the circle passing through the point ( x₁,y₁) with radius r is given by,

(x - x₁)² + (y - y₁)² = r²

To find the radius of circle

r =√[ (1 --3)² + (2 --1)²]

=√(4² + 3²)

= √(16 + 9)

=√25 = 5

To find the equation of the circle

(x₁, x₁) = (-3, -1) and r = 5

(x - x₁)² + (y - y₁)² = r²

(x - -3)² + (y - -1)² = 5²

(x + 3)² + (y +1)² = 25