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

2 Answers
Mar 23, 2018

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

Mar 23, 2018

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)