IDNLearn.com is designed to help you find reliable answers quickly and easily. Get accurate and comprehensive answers from our network of experienced professionals.

What is the radius of a circle whose equation is [tex]$x^2 + y^2 + 8x - 6y + 21 = 0$[/tex]?

A. 2 units
B. 3 units
C. 4 units
D. 5 units


Sagot :

To find the radius of the circle given by the equation [tex]\( x^2 + y^2 + 8x - 6y + 21 = 0 \)[/tex], we need to convert it into the standard form of a circle's equation, which is [tex]\( (x-h)^2 + (y-k)^2 = r^2 \)[/tex].

1. Reorganize the terms:
Group the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms together and move the constant term to the other side of the equation:
[tex]\[ x^2 + 8x + y^2 - 6y = -21 \][/tex]

2. Complete the square for the [tex]\( x \)[/tex] terms:
- Take the coefficient of [tex]\( x \)[/tex], which is 8, divide it by 2 to get 4, and then square it to get 16.
- Add and subtract this square inside the equation:
[tex]\[ x^2 + 8x + 16 + y^2 - 6y = -21 + 16 \][/tex]

3. Complete the square for the [tex]\( y \)[/tex] terms:
- Take the coefficient of [tex]\( y \)[/tex], which is -6, divide it by 2 to get -3, and then square it to get 9.
- Add and subtract this square inside the equation:
[tex]\[ x^2 + 8x + 16 + y^2 - 6y + 9 = -21 + 16 + 9 \][/tex]

4. Write both groups as perfect squares:
- The [tex]\( x \)[/tex] terms complete to [tex]\( (x + 4)^2 \)[/tex]
- The [tex]\( y \)[/tex] terms complete to [tex]\( (y - 3)^2 \)[/tex]
- The right side simplifies to 4:
[tex]\[ (x + 4)^2 + (y - 3)^2 = 4 \][/tex]

In the final equation, [tex]\( (x + 4)^2 + (y - 3)^2 = 4 \)[/tex], the standard form of a circle [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], the radius [tex]\( r \)[/tex] is the square root of the constant on the right-hand side.

So, the radius [tex]\( r \)[/tex] is [tex]\( \sqrt{4} = 2 \)[/tex] units.

Thus, the radius of the circle is 2 units.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.