Question

Find the center and radius of the circle with the given equation. Then select the correct
graph of the circle.
x2 + 8x + y2 - 14y + 40 = 0
The center of the circle is ___
and the radius is ___

Answer 1

Answer:

The center of the circle is $(-4, 7)$.

Its radius is $5$.

Step-by-step explanation:

Let $(a,\, b)$ represent the center of the circle, and let $r$ represent the radius of this circle. The equation for the circle would be:

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

Expand the left-hand side of this equation using the binomial expansion:

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

$x^2 + (- 2\,a)\, x + y^2 + (-2\, b)\, y + \left(a^2 + b^2 - r^2\right)= 0$.

Compare this equation with the one given in the question:

$x^2 + 8x + y^2 - 14\, y + 40 = 0$.

These two equations should represent the same circle. For that to happen, the coefficients should match. That gives a few equations about $a$, $b$, and $r$.

• The coefficients for $x$ should match: $(-2\,a ) =8 \implies a = -4$.
• The coefficients for $y$ should match: $(-2\, b) = -14 \implies b = 7$.
• The constant terms should match: $a^2 + b^2 - r^2 = 40$.

However, it is already determined that $a = -4$ and $b = 7$. That means $a^2 + b^2 - r^2 = (-4)^2 + 7^2 - r^2 = 65 - r^2$.

$65 - r^2 = 40 \implies r^2 = 25 \implies r = 5$.

Hence, the radius of this circle is $r = 5$. Also, its center is at $(a,\, b) = (-4,\, 7)$.