# Find the center and radius of the circle whose equation is given by: x^2 + y^2 + 6x - 10 y = 9

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### 3 Answers

The equation x^2 + y^2 + 6x - 10 y = 9 can be written as

x^2 + y^2 + 6x - 10 y = 9

=> x^2 + 6x + 9 + y^2 - 10y + 25 = 9 + 9 + 25

=> (x + 3)^2 + (y - 5)^2 = 43

=> (x + 3)^2 + (y - 5)^2 = (sqrt 43)^2

**This gives the radius of the circle as (-3 , 5) and the radius is sqrt 43**

x^2 + y^2 + 6x -10y = 9

To find the radius and the center we need to rewrite the equation into the standard form.

==> (x-a)^2 + (y-b)^2 = r^2 such that the center is (a,b) and the radius is r.

We will complete the square for x^2 and y^2.

==> x^2 + 6x + y^2 -10y = 9

==> (x^2 + 6x +9)-9 + (y^2 -10y +25) -25 = 9

==> (x+3)^2 + (y-5)^2 = 18+25

==> (x+3)^2 + (y-5)^2 = 43

**==> Center = (-3,5) and radius = sqrt43.**

We'll use the standard form of the equation of the circle:

(x-m)^2 + (y-n)^2 = r^2

(m,n) are the coordinates of the center of the circle and r represents the radius.

We'll use the technique of completing the squares in x and y:

x^2 + 6x + 9 + y^2 - 10y + 25 = 9+34

(x+3)^2 + (y-5)^2 = 43

**The coordinates of the center of the circle are (-3 ; 5) and the radius is r = sqrt 43.**