Questions: Find the center and radius of the circle whose equation is (x^2+x+y^2-7 y-4=0). The center of the circle is ( , ). The radius of the circle is .

Transcript text: Find the center and radius of the circle whose equation is $x^{2}+x+y^{2}-7 y-4=0$. The center of the circle is $\square$ , $\square$ ). The radius of the circle is $\square$ .

Solution

Solution Steps

Step 1: Rewrite the equation in standard form

The given equation of the circle is: \[ x^{2} + x + y^{2} - 7y - 4 = 0 \] To find the center and radius, we need to rewrite this equation in the standard form of a circle: \[ (x - h)^{2} + (y - k)^{2} = r^{2} \] where $(h, k)$ is the center and $r$ is the radius.

Step 2: Complete the square for $x$ and $y$

First, group the $x$ and $y$ terms: \[ x^{2} + x + y^{2} - 7y = 4 \]

Now, complete the square for the $x$ terms: \[ x^{2} + x = \left(x^{2} + x + \frac{1}{4}\right) - \frac{1}{4} = \left(x + \frac{1}{2}\right)^{2} - \frac{1}{4} \]

Next, complete the square for the $y$ terms: \[ y^{2} - 7y = \left(y^{2} - 7y + \frac{49}{4}\right) - \frac{49}{4} = \left(y - \frac{7}{2}\right)^{2} - \frac{49}{4} \]

Step 3: Substitute back into the equation

Substitute the completed squares back into the equation: \[ \left(x + \frac{1}{2}\right)^{2} - \frac{1}{4} + \left(y - \frac{7}{2}\right)^{2} - \frac{49}{4} = 4 \]

Combine the constants on the right side: \[ \left(x + \frac{1}{2}\right)^{2} + \left(y - \frac{7}{2}\right)^{2} = 4 + \frac{1}{4} + \frac{49}{4} \] \[ \left(x + \frac{1}{2}\right)^{2} + \left(y - \frac{7}{2}\right)^{2} = 4 + \frac{50}{4} = 4 + 12.5 = 16.5 \]

Step 4: Identify the center and radius

The equation is now in the standard form: \[ \left(x + \frac{1}{2}\right)^{2} + \left(y - \frac{7}{2}\right)^{2} = 16.5 \] From this, we can see that:

The center of the circle is $\left(-\frac{1}{2}, \frac{7}{2}\right)$.
The radius of the circle is $\sqrt{16.5} = \sqrt{\frac{33}{2}} = \frac{\sqrt{66}}{2}$.