Find the coordinate of the centre and radius of the following circle.
$x^{2} + y^{2} + 6 x - 8 y - 24 = 0$ .

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Solution

An equation in the form:
$(x - a)^{2} + (y - b)^{2} = r^{2}$
is the standard form for the equation of a circle with centre $(a, b)$ and radius r.
Lets try to convert the given equation:
$x^{2} + y^{2} + 6 x - 8 y - 24 = 0$
into the standard form for the equation of a circle.
Group the x and y terms separately and "move" the constant to the right side of the equation:
$x^{2} + 6 x + y^{2} - 8 y = 24$
Complete the square for each of x and y
$x^{2} + 6 x + 3^{2} + y^{2} - 8 y + 4^{2} = 24 + 3^{2} + 4^{2}$
Write the left side as the sum of two squared binomials and simplify the result on the right side
$(x + 3)^{2} + (y - 4)^{2} = 49$
Express the right side as a square.
$(x + 3)^{2} + (y - 4)^{2} = 7^{2}$
The equation for a circle with centre $(- 3, 4)$ and radius $7$ .

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Find the coordinate of the centre and radius of the following circle.x2+y2+6x−8y−24=0.

Find the coordinate of the centre and radius of the following circle.
$x^{2} + y^{2} + 6 x - 8 y - 24 = 0$ .