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SVETLANKA909090 [29]
3 years ago
8

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 .

Mathematics
2 answers:
Readme [11.4K]3 years ago
4 0

Answer:

(-21,-19)

\sqrt{849}

Standard form

Step-by-step explanation:

We are given the equation of circle

x^2+y^2+42x+38y-47=0

General equation of circle:

x^2+y^2+2gx+2fy+c=0

Centre: (-g,-f)

Radius: \sqrt{g^2+f^2-c}

Compare the equation to find f, g and c from the equation

g\rightarrow 21

f\rightarrow 19

c\rightarrow -47

Centre: (-21,-19)

Radius (r) =\sqrt{21^2+19^2+47}=\sqrt{849}

Standard form of circle:

(x+21)^2+(y+19)^2=849

The centre of circle at the point (-21,-19) and its radius is \sqrt{849}.

The general form of the equation of a circle that has the same radius as the above circle is standard form.

vesna_86 [32]3 years ago
3 0

Answer: The center of the circle is at the point (-21,-19), and its radius is  \sqrt{849} units.

The standard form of circle: (x+21)^2+(y+19)^2=849

Step-by-step explanation:

The given the equation of circle :-

x^2 + y^2 + 42x + 38y- 47 = 0

The general equation of circle is given by :

x^2+y^2+2hx+2gy+a=0

Here, the center of circle= (-h,-g)

and radius =\sqrt{h^2+g^2-a}

Now, Comparing the given equation to the general equation, we get  

2h=42\\\Rightarrow\ h=21\\\\2g=38\\\Rightarrow\ g=19

Thus, the center of the given circle= (-21,-19)

Radius of given circle = \sqrt{21^2+19^2+47}=\sqrt{849}

Now, the standard form of circle will be

(x-(-21))^2+(y-(-19))^2=(\sqrt{849})^2\\\\\Rightarrow\ (x+21)^2+(y+19)^2=849

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