Answer to Question #114121 in Analytic Geometry for Randal Rodriguez

Question

Given the equation of a circle x
2
+ y2
– 14x – 16y + 88 = 0. Determine the following
a. Center of the circle
b. Radius
c. Area
d. Perimeter

Answer 1

The given equation of circle is $x^2 + y^2 – 14x – 16y + 88 = 0$ . Let us rewrite it in a form $(x-x_0)^2 + (y-y_0)^2 = R^2, \quad (1)$

where $x_0$ and $y_0$ are the coordinates of the center of the circle.

If we expand the brackets in (1) and subtract $R^2$ from the both sides, we'll get

$x^2-2xx_0 + x_0^2 + y^2 - 2yy_0 + y_0^2 - R^2 = 0.$

Next we compare this formula and the given equation. We can see that $-2xx_0 = -14x, \; -2yy_0 = -16y,$

so $x_0 = 7, \;\; y_0 = 8.$

Therefore, $x_0^2+y_0^2 = 113.$ From (1) and the given equation we get

$x_0^2+y_0^2 - R^2 = 88$.

Therefore, $113-R^2 = 88, \;\; R^2 = 25, \;\; R=5.$

a) Center of the circle is (7, 8)

b) Radius is 5.

c) The area of the circle can be calculated as $S=\pi R^2 = \pi\cdot 5^2 = 25\pi \approx 78.54.$

d) Perimeter of the circle can be calculated as $P=2\pi R = 2\pi \cdot5 = 10\pi \approx 31.42.$