Answer:
The center-radius form of the equation of the circle is (x+4)^2+(y-1)^2=9^2(x+4)
2
+(y−1)
2
=9
2
.
Step-by-step explanation:
In this problem, the equation is in the general form of the equation of the circle. With that being said, we can't immediately determine the center and the radius. The equation has to be converted to center-radius form or standard form of the equation of the circle.
Equations of the Circle
General Form: x^2+y^2+Dx+Ey+F=0x
2
+y
2
+Dx+Ey+F=0
Standard Form or Center-Radius Form: (x-h)^2+(y-k)^2=r^2(x−h)
2
+(y−k)
2
=r
2
Given equation is x^2+y^2+8x-2y-64=0x
2
+y
2
+8x−2y−64=0 .
Step by Step Process
1. Arrange the terms. Put together all the similar terms and isolate the constant.
\begin{gathered}\begin{aligned}x^2+y^2+8x-2y-64&=0\\(x^2+8x)+(y^2-2y)&=64\end{aligned}\end{gathered}
x
2
+y
2
+8x−2y−64
(x
2
+8x)+(y
2
−2y)
=0
=64
2. The equation in the form (ax^2+b_1x)+(ay^2+b_2y)=c(ax
2
+b
1
x)+(ay
2
+b
2
y)=c . Complete the squares by adding \left(\frac{b_1}{2}\right)^2=\left(\frac{8}{2}\right)^2=16(
2
b
1
)
2
=(
2
8
)
2
=16 and \left(\frac{b_2}{2}\right)^2=\left(\frac{-2}{2}\right)^2=1(
2
b
2
)
2
=(
2
−2
)
2
=1 on both sides of the equation.
\begin{gathered}\begin{aligned}(x^2+8x)+(y^2-2y)&=64\\(x^2+8x+16)+(y^2-2y+1)&=64+16+1\\(x^2+8x+16)+(y^2-2y+1)&=81\end{aligned}\end{gathered}
(x
2
+8x)+(y
2
−2y)
(x
2
+8x+16)+(y
2
−2y+1)
(x
2
+8x+16)+(y
2
−2y+1)
=64
=64+16+1
=81
3. Factor the perfect square trinomials.
\begin{gathered}\begin{aligned}(x^2+8x+16)+(y^2-2y+1)&=81\\(x+4)^2+(y-1)^2=9^2\end{aligned}\end{gathered}
(x
2
+8x+16)+(y
2
−2y+1)
(x+4)
2
+(y−1)
2
=9
2
=81
Thus, the enter-radius form of the equation of the circle is (x+4)^2+(y-1)^2=9^2(x+4)
2
+(y−1)
2
=9
2
. The radius is 9 units and the center is at (-4, 1)(−4,1) .
To learn more about the equation of the circle, go to
