How do you find the center and radius of the circle: # x^2 + y^2 – 10x + 6y + 18 = 0#?
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Centre is #(5,-3)# and the Radius is #4#
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
centre : #C(5,-3)#
radius : #r=4#
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)#
