Question

The equation of a circle is x2 + y2 + 12x 10y + 52 = 0.

Part A
What are the coordinates of the center point of the circle?
Enter a number in each box.
Part B
7 =
What is the radius of the circle?
Enter a number in the box.

Answer 1

center: (-6,-5)

radius =
`√(113)`

Explanation:

completing the square to factor

x2+12x + __ + y2 + 10y + __ = 52

__ = 36 __ = 25

x2+12x+36 + y2+10y+25 = 52+36+25

(x+6)^2 + (y+5)^2 = 113

center: (-6,-5)

radius =
`√(113)`

Answer 2

A: (-6,-5)

B: r = 3

Explanation:

To find the center and radius of the circle given by the equation x^2 + y^2 + 12x + 10y + 52 = 0, we first need to rewrite the equation in standard form, which is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is its radius.

Part A:

To rewrite the equation in standard form, we need to complete the square for both x and y. We can do this by adding and subtracting the square of half the coefficient of x and y respectively, as follows:

x^2 + 12x + y^2 + 10y + 52 = 0

(x^2 + 12x + 36) + (y^2 + 10y + 25) = -52 + 36 + 25

(x + 6)^2 + (y + 5)^2 = 9

Now we can see that the equation is in standard form, with center (-6, -5) and radius 3. Therefore, the center point of the circle is (-6, -5).

Part B:

The radius of the circle is given by the square root of the right-hand side of the standard form equation, which is 3. Therefore, the radius of the circle is 3.