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Someone help me, please

find the standard equation of a circle given the following equation: x^2+4x+y^2-6y-3=0

User Red Fx
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1 Answer

4 votes

Answer:

( x+2) ^2 +(y-3)^2=4^2

The center is ( -2,3) and the radius is 4

Explanation:

x^2+4x+y^2-6y-3=0

Add 3 to each side

x^2+4x+y^2-6y=3

Complete the square for x

4/2 = 2 2^2 = 4 so add 4 to each side

x^2+ 4x +4+y^2-6y=3+4

( x+2) ^2 +y^2-6y=7

Complete the square for y

-6/2 = -3 (-3) ^2 = 9 so add 9 to each side

( x+2) ^2 +y^2-6y +9=7+9

( x+2) ^2 +(y-3)^2=16

( x+2) ^2 +(y-3)^2=4^2

The standard equation of a circle is

(x – h)^2 + (y – k)^2 = r^2 where ( h,k) is the center and r is the radius

User Markau
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