Question

How do you find the center and radius of the circle: `x^2 + y^2 – 10x + 6y + 18 = 0`?

Answer 1

Centre is `(5,-3)` and the Radius is `4`

Explanation:

We must write this equation in the form `(x-a)^2 + (y-b)^2 = r^2`
Where `(a,b)` are the co ordinates of the center of the circle and the radius is `r`.

So the equation is `x^2 + y^2 -10x + 6y +18 = 0`

Complete the squares so add 25 on both sides of the equation

`x^2 + y^2 -10x + 25 + 6y +18 = 0+25`

= `(x-5)^2 + y^2+ 6y +18 = 0+25`

Now add 9 on both sides

`(x-5)^2 + y^2+ 6y +18 +9= 0+25+9`
=`(x-5)^2 + (y+3)^2 +18 = 0+25+9`

This becomes

`(x-5)^2 + (y+3)^2 = 16`

So we can see that the centre is `(5,-3)` and the radius is `sqrt(16)` or 4

Answer 2

centre : `C(5,-3)`
radius : `r=4`

Explanation:

The general equation of a circle:

`color(red)(x^2+y^2+2gx+2fy+c=0...........to(1)`,

whose centre is `color(red)(C((-g,-f))` and radius is `color(red)(r=sqrt(g^2+f^2-c)`

We have,

`x^2+y^2-10x+6y+18=0`

Comparing with `equ^n(1)`, we get

`2g=-10,2f=6 and c=18`

`=>g=-5,f=3 and c=18`

So,

radius `r=sqrt((-5)^2+(3)^2-18)=sqrt(25+9-18)=sqrt(16)=4`

i.e. `r=4>0`

centre `C(-g,-f)=>C(-(-5),-3)`

i.e. centre `C(5,-3)`