Question

NO LINKS!! Find the center and radius of the circle with the given equation​

Answer 1

`\textsf{center \quad $(x,y)=\left(\; \boxed{6,-2}\; \right)$}`

`\textsf{radius \quad $r=\boxed{8}$}`

Explanation:

`\boxed{\begin{minipage}{4 cm}\underline{Equation of a circle}\\\\$(x-a)^2+(y-b)^2=r^2$\\\\where:\\ \phantom{ww}$\bullet$ $(a, b)$ is the center. \\ \phantom{ww}$\bullet$ $r$ is the radius.\\\end{minipage}}`

Given equation:

`x^2+y^2-12x+4y-24=0`

Move the constant to the right side of the equation and collect like variables on the left side of the equation:

`\implies x^2-12x+y^2+4y=24`

Create perfect square trinomials for the both variables by adding the square of the half the coefficient of x and y to both sides:

`\implies x^2-12x+\left((-12)/(2)\right)^2+y^2+4y+\left((4)/(2)\right)^2=24+\left((-12)/(2)\right)^2+\left((4)/(2)\right)^2`

`\implies x^2-12x+\left(-6\right)^2+y^2+4y+\left(2\right)^2=24+\left(-6\right)^2+\left(2\right)^2`

`\implies x^2-12x+36+y^2+4y+4=24+36+4`

`\implies x^2-12x+36+y^2+4y+4=64`

Factor the two perfect trinomials:

`\implies (x-6)^2+(y+2)^2=64`

Therefore:

`\textsf{center \quad $(x,y)=\left(\; \boxed{6,-2}\; \right)$}`

`\textsf{radius \quad $r=\boxed{8}$}`