Question

The equation of a circle is x2 + y2 + 6x + 4y + 10 = 1. What is this equation written in its standard form?

Answer 1

Option A: $(x+3)^2+(y+2)^2=4$ is the equation of the circle

Explanation:

The equation of the circle is $x^{2} +y^2+6x+4y+10=1$

We need to write the equation of the circle in standard form.

Let us subtract 10 from both sides of the equation.

Thus, we have,

$x^{2} +y^2+6x+4y=-9$

Grouping the terms,we have,

$(x^{2} +6x)+(y^2+4y)=-9$

Now, we shall complete the square, let us add and subtract the equation by 9 to write the term $\left(x^{2}+6 x\right)$ in the form of $(a+b)^2$

Thus, we have,

$(x^{2} +6x+9-9)+(y^2+4y)=-9$

Simplifying, we get,

$(x^{2} +6x+9)+(y^2+4y)=-9+9$

       $(x+3)^2+(y^2+4y)=0$

Similarly, we shall complete the square for the term $\left(y^{2}+4 y\right)$ in the form of $(a+b)^2$

Thus, we have,

$(x+3)^2+(y^2+4y+4-4)=0$

Simplifying, we get,

$(x+3)^2+(y^2+4y+4)=4$

       $(x+3)^2+(y+2)^2=4$

Thus, the equation of the circle in standard form is $(x+3)^2+(y+2)^2=4$

Hence, Option A is the correct answer.