Respuesta :
Answer:
[tex](x+2)^2+(y-3)^2=9[/tex]• The center, (a, b) = (-2, 3)
,• Radius = 3
Explanation:
Given the equation of the circle below:
[tex]x^2+4x+y^2−6y=−4[/tex]We are required to complete the square, then identify the center and radius of the circle.
First, to complete the square for x, divide the coefficient of x by 2, square it and add it to both sides of the equation.
[tex]x^2+4x+\left(\frac{4}{2}\right)^2+y^2−6y=−4+\left(\frac{4}{2}\right)^2[/tex]Repeat the same process for y.
[tex]x^2+4x+\left(\frac{4}{2}\right)^2+y^2−6y+\left(-\frac{6}{2}\right)^2=−4+\left(\frac{4}{2}\right)^2+\left(-\frac{6}{2}\right)^2[/tex]Write the left side as perfect squares of x and y and simplify the right side:
[tex]\begin{gathered} x^2+4x+\left(2\right)^2+y^2−6y+\left(-3\right)^2=−4+\left(2\right)^2+\left(-3\frac{}{}\right)^2 \\ (x+2)^2+(y-3)^2=-4+4+9 \\ (x+2)^2+(y-3)^2=9 \end{gathered}[/tex]The equation rewritten is:
[tex](x+2)^2+(y-3)^2=9[/tex]Compared with the standard equation of a circle:
[tex](x-a)^2+(y-b)^2=r^2[/tex]• The center, (a, b) = (-2, 3)
,• Radius = 3
