Question

What is the center and radius of x^2+6x+y^2-4y =-4?​

Answer 1

Answer:

The center of this circle is at $(-3,\, 2)$. The radius of this circle is $3$.

Step-by-step explanation:

A circle with center $(a,\, b)$ and a radius of $r$ ($r > 0$) could be expressed as:

$(x - a)^{2} + (y - b)^{2} = r^{2}$.

(In other words, a point is on this circle if and only if the distance between that point and the center is equal to the radius.)

Rearrange this equation using binomial expansion to match the equation given in this question:

$(x^{2} - 2\, a\, x + a^{2}) + (y^{2} - 2\, b\, y + b^{2}) = r^{2}$.

$x^{2} + (-2\, a)\, x + y^{2} + (-2\, b)\, y = -a^{2} - b^{2} + r^{2}$.

The equation in this question is:

$x^{2} + 6\, x + y^{2} + (-4)\, y = -4$.

Match up the coefficients of $x$ and $y$ in the two equations.:

• Coefficient of $x$: $(-2\, a) = 6$.
• Coefficient of $y$: $(-2\, b) = (-4)$.

Thus, $a = (-3)$ and $b = 2$.

The constants of the two equations should also match up:

$(-a^{2} - b^{2} + r^{2}) = (-4)$.

Substitute in $a = (-3)$ as well as $b = 2$ and solve for $r$:

$r^{2} = 9$.

$r = 3$ (since $r > 0$.)

Therefore, the center of this circle is $(-3,\, 2)$. The radius of this circle is $3$.