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Math  /  Geometry

QuestionQuestion 15 of 40 What are the center and radius of the circle defined by the equation x2+y2−6x+10y+25=0x^{2}+y^{2}-6 x+10 y+25=0 ? A. Center (−3,5)(-3,5); radius 9 B. Center (3,−5)(3,-5); radius 3 C. Center (3,−5)(3,-5); radius 9 D. Center (−3,5)(-3,5); radius 3 SUBMIT

Studdy Solution

STEP 1

1. The given equation is of a circle in the general form.
2. The equation needs to be converted to the standard form of a circle.

STEP 2

1. Rewrite the equation in the standard form of a circle.
2. Identify the center of the circle.
3. Identify the radius of the circle.

STEP 3

Rewrite the given equation in the standard form of a circle by completing the square for both x x and y y .
The given equation is:
x2+y2−6x+10y+25=0 x^2 + y^2 - 6x + 10y + 25 = 0
First, group the x x and y y terms:
(x2−6x)+(y2+10y)=−25 (x^2 - 6x) + (y^2 + 10y) = -25
Complete the square for x x :
x2−6x x^2 - 6x
Take half of the coefficient of x x , square it, and add/subtract it:
x2−6x+9−9 x^2 - 6x + 9 - 9
This becomes:
(x−3)2−9 (x - 3)^2 - 9
Complete the square for y y :
y2+10y y^2 + 10y
Take half of the coefficient of y y , square it, and add/subtract it:
y2+10y+25−25 y^2 + 10y + 25 - 25
This becomes:
(y+5)2−25 (y + 5)^2 - 25
Substitute back into the equation:
(x−3)2−9+(y+5)2−25=−25 (x - 3)^2 - 9 + (y + 5)^2 - 25 = -25
Simplify:
(x−3)2+(y+5)2=9 (x - 3)^2 + (y + 5)^2 = 9

STEP 4

Identify the center of the circle from the standard form equation:
The standard form of a circle is:
(x−h)2+(y−k)2=r2 (x - h)^2 + (y - k)^2 = r^2
From the equation:
(x−3)2+(y+5)2=9 (x - 3)^2 + (y + 5)^2 = 9
The center (h,k)(h, k) is:
(3,−5) (3, -5)

STEP 5

Identify the radius of the circle:
The radius r r is the square root of the right side of the equation:
r2=9 r^2 = 9
r=9=3 r = \sqrt{9} = 3
The center and radius of the circle are:
Center: (3,−5)(3, -5), Radius: 33
The correct answer is:
B. Center (3,−5)(3, -5); radius 3

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