What is the radius of a circle given by the equation [tex]$x^2 + y^2 - 2x + 8y - 47 = 0$[/tex]?

Radius [tex]=$\square$[/tex] units



Answer :

To determine the radius of the circle given by the equation [tex]\(x^2 + y^2 - 2x + 8y - 47 = 0\)[/tex], we need to rewrite the equation in the standard form of a circle's equation, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is its radius.

We can achieve this by completing the square for both the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.

1. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together:
[tex]\[ x^2 - 2x + y^2 + 8y = 47 \][/tex]

2. Complete the square for the [tex]\(x\)[/tex] terms:
[tex]\[ x^2 - 2x \][/tex]
To complete the square:
- Take the coefficient of [tex]\(x\)[/tex], which is [tex]\(-2\)[/tex], divide by 2 and square it: [tex]\(\left(\frac{-2}{2}\right)^2 = 1\)[/tex].
- Add and subtract this square inside the equation:
[tex]\[ x^2 - 2x + 1 - 1 = (x - 1)^2 - 1 \][/tex]

3. Complete the square for the [tex]\(y\)[/tex] terms:
[tex]\[ y^2 + 8y \][/tex]
To complete the square:
- Take the coefficient of [tex]\(y\)[/tex], which is [tex]\(8\)[/tex], divide by 2 and square it: [tex]\(\left(\frac{8}{2}\right)^2 = 16\)[/tex].
- Add and subtract this square inside the equation:
[tex]\[ y^2 + 8y + 16 - 16 = (y + 4)^2 - 16 \][/tex]

4. Substitute the completed squares back into the equation:
[tex]\[ (x - 1)^2 - 1 + (y + 4)^2 - 16 = 47 \][/tex]

5. Combine and simplify the constants on the right side:
[tex]\[ (x - 1)^2 + (y + 4)^2 - 17 = 47 \][/tex]
[tex]\[ (x - 1)^2 + (y + 4)^2 = 47 + 17 \][/tex]
[tex]\[ (x - 1)^2 + (y + 4)^2 = 64 \][/tex]

Now we have the equation in the standard form:
[tex]\[ (x - 1)^2 + (y + 4)^2 = 64 \][/tex]

From this, we can see that the radius [tex]\(r\)[/tex] of the circle is:
[tex]\[ r^2 = 64 \implies r = \sqrt{64} = 8 \][/tex]

Thus, the radius of the circle is:
[tex]\[ \boxed{8} \text{ units} \][/tex]

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