Question

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 .

Answer 1

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.

Answer 2

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

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