Need help finding (h,K,) and the radius of a circle

**M670** · October 20th 2012, 08:38 AM

so my equation is x²+y²-4x+2y-4=0
i know the formula is (x-h)²+(y-k)²=r²

x²-4x+y²+2y-4=0

the Y is (y-2)² which makes my k equal to positive 2 just having trouble with the h and radius

**skeeter** · October 20th 2012, 08:50 AM

$x^2+y^2-4x+2y-4 = 0$

$x^2 - 4x + y^2 + 2y = 4$

complete the square for x and y ...

$(x^2 - 4x + 4) + (y^2 + 2y + 1) = 4 + 4 + 1$

finish it ...

**Soroban** · October 20th 2012, 09:03 AM

Hello, M670!

Sorry, your algebra is off . . .

$\text{Find the center and radius of the circle: }\:x^2+y^2-4x+2y-4\:=\:0$

We must "complete the square".

$\begin{array}{c}x^2 + y^2 - 4x + 2y - 4 \:=\:0 \\ \\ x^2 - 4x + y^2 + 2y \:=\:4 \\ \\ x^2 - 4x \,{\color{red}+\, 4} + y^2 + 2y {\color{blue}\,+\, 1} \:=\:4{\color{red}\,+\,4}{\color{blue}\,+\,1} \\ \\ (x-2)^2 +(y+1)^2 \:=\:9 \end{array}$

The circle has center $(2,\text{-}1)$ and radius $r = 3.$

**M670** · October 20th 2012, 09:33 AM

ok I understand that question but given this example here 5 x^2 - 10 x +5 y^2 - 8 y - 6 = 0 .

I worked it down to (x-1)^2 and (y-0.8)^2 =
I divided out 5 and was left with x^2-2x+y^2-1.6y-1.2=0
then its x^2-2x+1 + y^2-1.6y ?= 1.2-1 (?)

**Plato** · October 20th 2012, 10:01 AM

Originally Posted by M670

ok I understand that question but given this example here 5 x^2 - 10 x +5 y^2 - 8 y - 6 = 0 .

I worked it down to (x-1)^2 and (y-0.8)^2 =
I divided out 5 and was left with x^2-2x+y^2-1.6y-1.2=0
then its x^2-2x+1 + y^2-1.6y ?= 1.2-1 (?)

Why do you have $1.2-1 (?)$ ?
It clearly should be $1.2+(1)+(.64)$

**M670** · October 20th 2012, 10:13 AM

Originally Posted by Plato

Why do you have $1.2-1 (?)$ ?
It clearly should be $1.2+(1)+(.64)$

The ? was because I was unsure if it would be .64 or something esle

**M670** · October 20th 2012, 10:15 AM

How do you figure it out when you have a fraction or decimal to complete the squares?

**Plato** · October 20th 2012, 10:23 AM

Originally Posted by M670

How do you figure it out when you have a fraction or decimal to complete the squares?

It is one-half the coefficient of the 'middle term' squared.
$\left(\frac{-1.6}{2}\right)^2=0.64$ .

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