Step 1
we know that
The equation of a circle in standard form is equal to
[tex](x-h)^{2} +(y-k)^{2}=r^{2}[/tex]
where
(h,k) is the center of the circle
r is the radius of the circle
In this problem we have
[tex]x^{2} +y^{2} +42x+38y-47=0[/tex]
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}+42x)+(y^{2}+38y)=47[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=47+21^{2}+19^{2}[/tex]
[tex](x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=849[/tex]
Rewrite as perfect squares
[tex](x+21)^{2}+(y+19)^{2}=849[/tex]
The center of the circle is the point [tex](-21,-19)[/tex]
The radius of the circle is [tex]\sqrt{849}\ units[/tex]
The answer Part a) is
The equation of the circle in standard form is equal to
[tex](x+21)^{2}+(y+19)^{2}=849[/tex]
The answer Part b) is
The center of the circle is the point [tex](-21,-19)[/tex]
The answer Part c) is
The radius of the circle is [tex]\sqrt{849}\ units[/tex]
Let's verify each case to determine the solution of the second part of the problem
Step 2
we have
[tex]x^{2} +y^{2} +60x+14y+98=0[/tex]
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}+60x)+(y^{2}+14y)=-98[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=-98+30^{2}+7^{2}[/tex]
[tex](x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=851[/tex]
Rewrite as perfect squares
[tex](x+30)^{2}+(y+7)^{2}=851[/tex]
The radius of the circle is [tex]\sqrt{851}\ units[/tex]
therefore
This circle does not have the same radius of the circle above
Step 3
we have
[tex]x^{2} +y^{2} +44x-44y+117=0[/tex]
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}+44x)+(y^{2}-44y)=-117[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=-117+22^{2}+22^{2}[/tex]
[tex](x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=851[/tex]
Rewrite as perfect squares
[tex](x+22)^{2}+(y-22)^{2}=851[/tex]
The radius of the circle is [tex]\sqrt{851}\ units[/tex]
therefore
This circle does not have the same radius of the circle above
Step 4
we have
[tex]x^{2} +y^{2} -38x+42y+74=0[/tex]
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}-38x)+(y^{2}+42y)=-74[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=-74+19^{2}+21^{2}[/tex]
[tex](x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=728[/tex]
Rewrite as perfect squares
[tex](x-19)^{2}+(y+21)^{2}=728[/tex]
The radius of the circle is [tex]\sqrt{728}\ units[/tex]
therefore
This circle does not have the same radius of the circle above
Step 5
we have
[tex]x^{2} +y^{2} -50x-30y+1=0[/tex]
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}-50x)+(y^{2}-30y)=-1[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}-50x+25^{2})+(y^{2}-30y+15^{2})=-1+25^{2}+15^{2}[/tex]
[tex](x^{2}-50x+25^{2})+(y^{2}-30y+15^{2})=849[/tex]
Rewrite as perfect squares
[tex](x-25)^{2}+(y-15)^{2}=849[/tex]
The radius of the circle is [tex]\sqrt{849}\ units[/tex]
therefore
This circle has the same radius of the circle above
therefore
The answer is
[tex]x^{2} +y^{2} -50x-30y+1=0[/tex] -----> has the same radio that the circle above
