Question

What is the general form of the equation of the given circle with center A ? ​

Answer 1

The equation for a circle is (x-h)^2+(y-k^2)=r^2. Where (h,k) represents the center and r is the radius. So, by substitution, the equation becomes (x+3)^2+(y-12)^2=5^2. This becomes x^2+6x+9+y^2-24y+144=25. Simplified to x^2+y^2+6x-24y+128=0. The answer is B.

Answer 2

the equation of circle is x² + y² + 6x - 24y + 128 = 0

Explanation:

To find the general form of the equation of the circle with center A (-3, 12) and radius r=5, we simply use this formula;

General Equation of a circle is (x - a)² + (y-b)² = r²

(a,b) are the two point at the center of the circle which are (-3, 12)

which implies a = -3 and b = 12

r is the radius of the circle which is given as 5 from the above diagram

to get r², we simply square 5 and so r² = 25

We can now plug in the values of our variables into the equation

(x - a)² + (y-b)² = r²

(x - [-3])² + ( y - 12)² = 5²

(x +3)² + ( y - 12)² = 5²

we will expand all the brackets

x² + 6x + 9 + y² -24y + 144 = 25

x² +6x + y² -24y + 153 = 25

Take 25 to the left hand side of the equation

x² +6x + y² -24y + 153 - 25 = 0

x² +6x + y² -24y + 128 = 0

Rearranging the equation to give us a standard form of the equation of the circle, we have;

x² + y² + 6x - 24y + 128 =0

{Note: (x + 3)² = (x+3)(x+3) = x² + 6x + 9 and

(y - 12)² = (y -12)(y-12) = y² - 24y + 144}

Therefore the general form of the equation of a circle with center(-3, 12) and radius 5 is x² + y² + 6x - 24y + 128 =0