Question

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

Answer 1

`\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{12}{ k})\qquad \qquad radius=\stackrel{5}{ r} \\\\\\\ [x-(-3)]^2+[y-12]^2=5^2\implies (x+3)^2+(y-12)^2=25`

Answer 2

The general form of the equation is x² + y² + 6x - 24y + 128 =0 .

Explanation:

The general form of the equation is written in the general form

(x - h)² + (y - k)² = r²

As given the figure in the question

Centre of the circle is A (-3 , 12) and radius is 5 .

Thus

(x - (-3))² + (y - 12)² = 5²

5² = 25

(x - (-3))² + (y - 12)² = 5²

(x + 3)² + (y - 12)² = 25

(By using the formula (a + b)² = a² + b² + 2ab , (a - b)² = a² + b² - 2ab)

Thus

x² + 3² + 2 × 3 × x + y² + 12² - 2 × 12 × y = 25

3² = 9

12² = 144

x² + 9 + 6x + y² + 144 - 24y = 25

x² + y² + 6x -24y + 9 + 144 - 25 = 0

x² + y² + 6x - 24y + 128 =0

Therefore the general form of the equation is x² + y² + 6x - 24y + 128 =0 .