Finding the Centre Point and Radius of a Circle

[chesterdg123]

First we need to put it into the equation of a circle's standard form



X^2=y^2-12x+8y+3=0

(x^2-12x)+(y^2+8y)=-3

(x-2?3x)^2 +(y+2?2x)=-3

(-2?3x, 2V2x) radius = ?-3

Does this look right?

Thanks!

 

[galactus]

I am sorry to say, no. You have a radius of \(\displaystyle \sqrt{-3}\). That is complex. How can you have an imaginary radius?.

I will show you this one so you know how. Okey-doke?.

Group and complete the square(which is what you failed to do).

\(\displaystyle x^{2}+y^{2}-12x+8y=-3\)

\(\displaystyle (x^{2}-12x+...)+(y^{2}+8y+...)=-3\)

The square of half the coefficient of x added to both sides:

\(\displaystyle (x^{2}-12x+36)+(y^{2}+8y+16)=-3+36+16\)

Factor:

\(\displaystyle (x-6)^{2}+(y+4)^{2}=49\)

The circle has center at (6,-4) and radius 7

Understand everything I done?. It not, let me know.