Question #114121
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
1
Expert's answer
2020-05-06T18:23:53-0400

The given equation of circle is x2+y214x16y+88=0x^2 + y^2 – 14x – 16y + 88 = 0 . Let us rewrite it in a form (xx0)2+(yy0)2=R2,(1)(x-x_0)^2 + (y-y_0)^2 = R^2, \quad (1)

where x0x_0 and y0y_0 are the coordinates of the center of the circle.

If we expand the brackets in (1) and subtract R2R^2 from the both sides, we'll get

x22xx0+x02+y22yy0+y02R2=0.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 2xx0=14x,  2yy0=16y,-2xx_0 = -14x, \; -2yy_0 = -16y,

so x0=7,    y0=8.x_0 = 7, \;\; y_0 = 8.

Therefore, x02+y02=113.x_0^2+y_0^2 = 113. From (1) and the given equation we get

x02+y02R2=88x_0^2+y_0^2 - R^2 = 88.

Therefore, 113R2=88,    R2=25,    R=5.113-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=πR2=π52=25π78.54.S=\pi R^2 = \pi\cdot 5^2 = 25\pi \approx 78.54.

d) Perimeter of the circle can be calculated as P=2πR=2π5=10π31.42.P=2\pi R = 2\pi \cdot5 = 10\pi \approx 31.42.


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