Mathematics, 13.01.2020 16:31 bryantjorell

The general form of the equation of a circle is x2 + y2 + 42x + 38y − 47 = 0. the equation of this circle in standard form is (x - 21)^2 + (y - 19)^2 = 127 (x + 21)^2 + (y + 19)^2 = 849 (x + 21)^2 + (y + 19)^2 = 851 (x - 19)^2 + (y - 21)^2 = 2,209 . the center of the circle is at the point (-19, -21) (-21, -19) (19, 21) (21, 19) , and its radius is 127^(1/2) 849^(1/2) 851^(1/2) 47 units. the general form of the equation of a circle that has the same radius as the above circle is x^2 + y^2 + 60x + 14y + 98 = 0 x^2 + y^2 + 44x - 44y + 117 = 0 x^2 + y^2 - 38x + 42y + 74 = 0 x^2 + y^2 - 50x - 30y + 1 = 0 .

Answers: 1

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Answer from: ddswaggsoflyy3176

Step 1

we know that

The equation of a 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

In this problem we have

$x^{2} +y^{2} +42x+38y-47=0$

Convert to standard form

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

$(x^{2}+42x)+(y^{2}+38y)=47$

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

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=47+21^{2}+19^{2}$

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=849$

Rewrite as perfect squares

$(x+21)^{2}+(y+19)^{2}=849$

The center of the circle is the point (-21,-19)

The radius of the circle is $\sqrt{849}\ units$

The answer Part a) is

The equation of the circle in standard form is equal to

$(x+21)^{2}+(y+19)^{2}=849$

The answer Part b) is

The center of the circle is the point (-21,-19)

The answer Part c) is

The radius of the circle is $\sqrt{849}\ units$

Let's verify each case to determine the solution of the second part of the problem

Step 2

we have

$x^{2} +y^{2} +60x+14y+98=0$

Convert to standard form

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

$(x^{2}+60x)+(y^{2}+14y)=-98$

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

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=-98+30^{2}+7^{2}$

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=851$

Rewrite as perfect squares

$(x+30)^{2}+(y+7)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 3

we have

$x^{2} +y^{2} +44x-44y+117=0$

Convert to standard form

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

$(x^{2}+44x)+(y^{2}-44y)=-117$

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

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=-117+22^{2}+22^{2}$

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=851$

Rewrite as perfect squares

$(x+22)^{2}+(y-22)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 4

we have

$x^{2} +y^{2} -38x+42y+74=0$

Convert to standard form

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

$(x^{2}-38x)+(y^{2}+42y)=-74$

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

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=-74+19^{2}+21^{2}$

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=728$

Rewrite as perfect squares

$(x-19)^{2}+(y+21)^{2}=728$

The radius of the circle is $\sqrt{728}\ units$

therefore

This circle does not have the same radius of the circle above

Step 5

we have

$x^{2} +y^{2} -50x-30y+1=0$

Convert to standard form

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

$(x^{2}-50x)+(y^{2}-30y)=-1$

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

$(x^{2}-50x+25^{2})+(y^{2}-30y+15^{2})=-1+25^{2}+15^{2}$

$(x^{2}-50x+25^{2})+(y^{2}-30y+15^{2})=849$

Rewrite as perfect squares

$(x-25)^{2}+(y-15)^{2}=849$

The radius of the circle is $\sqrt{849}\ units$

therefore

This circle has the same radius of the circle above

therefore

The answer is

$x^{2} +y^{2} -50x-30y+1=0$ -----> has the same radio that the circle above

Answer from: Quest

-14

step-by-step explanation:

2x2 – 2(x+4)

=2x2-2（5+4）

=4-2×9

=4-18

=-14

Answer from: Quest

may i see the picture? it would a lot, xox

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