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The general form of the equation of a circle is x2 + y2 + 8x + 22y + 37 = 0.

The equation of this circle in standard form is (x +

)2 + (y +

)2 =

. The center of the circle is at the point (

,

).

Verified Answer :

caylus

Hello,

x²+y²+8x+22y+37=0
==>(x²+24x+16)+(y²+211y+121)+37-16-121=0
==>(x+4)²+(y+1)²=10²
Center=(-4,-1) , radius=10

calculista

we know that

the equation of the circle in standard form is equal to

\((x-h)^{2}+(y-k)^{2}=r^{2}\)

where

\((h,k)\) is the center of the circle

r is the radius of the circle

we have

\(x^{2}+y^{2} +8x+22y+37=0\)

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

\((x^{2}+8x)+(y^{2} +22y)=-37\)

Complete the square twice. Remember to balance the equation by adding the same constants to each side

\((x^{2}+8x+4^{2})+(y^{2} +22y+11^{2})=-37+4^{2}+11^{2}\)

\((x^{2}+8x+16)+(y^{2} +22y+121)=-37+16+121\)

\((x^{2}+8x+16)+(y^{2} +22y+121)=100\)

Rewrite as perfect squares

\((x+4)^{2}+(y+11)^{2}=10^{2}\)

the  center is the point \((-4,-11)\)

the radius is \(r=10\ units\)

therefore

The answer is

a) the equation in standard form is \((x+4)^{2}+(y+11)^{2}=10^{2}\)

b) the  center of the circle is the point \((-4,-11)\) and the radius of the circle is \(r=10\ units\)

see the attached figure to better understand the problem

View image calculista

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