To find the $x$- and $y$-intercepts for the circle, let's analyze its equation:

$(x - 3)^2 + \left( y - \frac{3}{2} \right)^2 = \frac{\sqrt{85}}{2}$

However, there seems to be an inconsistency because the radius term on the right side should be the square of the radius, not the square root. Let's assume the radius squared should be $\frac{85}{2}$ (since that would make more sense in the context of a circle equation).

So the equation becomes:

$(x - 3)^2 + \left( y - \frac{3}{2} \right)^2 = \frac{85}{2}$

Step 1: Find the $y$-intercept

The $y$-intercept occurs when $x = 0$. Substitute $x = 0$ into the equation:

$(0 - 3)^2 + \left( y - \frac{3}{2} \right)^2 = \frac{85}{2}$

This simplifies to:

$9 + \left( y - \frac{3}{2} \right)^2 = \frac{85}{2}$

Subtract 9 from both sides (noting that $9 = \frac{18}{2}$):

$\left( y - \frac{3}{2} \right)^2 = \frac{85}{2} - \frac{18}{2} = \frac{67}{2}$

Now, take the square root of both sides:

$y - \frac{3}{2} = \pm \sqrt{\frac{67}{2}}$

Solve for $y$:

$y = \frac{3}{2} \pm \sqrt{\frac{67}{2}}$

So, the $y$-intercepts are:

$y = \frac{3}{2} + \sqrt{\frac{67}{2}} \quad \text{and} \quad y = \frac{3}{2} - \sqrt{\frac{67}{2}}$

Step 2: Find the $x$-intercept

The $x$-intercept occurs when $y = 0$. Substitute $y = 0$ into the equation:

$(x - 3)^2 + \left(0 - \frac{3}{2} \right)^2 = \frac{85}{2}$

Simplify the term for $y = 0$:

$(x - 3)^2 + \left(\frac{3}{2}\right)^2 = \frac{85}{2}$

Since $\left( \frac{3}{2} \right)^2 = \frac{9}{4}$, we rewrite the equation as:

$(x - 3)^2 + \frac{9}{4} = \frac{85}{2}$

Subtract $\frac{9}{4}$ from both sides. Convert $\frac{85}{2}$ to a fraction with a denominator of 4:

$(x - 3)^2 = \frac{170}{4} - \frac{9}{4} = \frac{161}{4}$

Now, take the square root of both sides:

$x - 3 = \pm \sqrt{\frac{161}{4}} = \pm \frac{\sqrt{161}}{2}$

Solve for $x$:

$x = 3 \pm \frac{\sqrt{161}}{2}$

So, the $x$-intercepts are:

$x = 3 + \frac{\sqrt{161}}{2} \quad \text{and} \quad x = 3 - \frac{\sqrt{161}}{2}$

Summary

• $y$-intercepts: $y = \frac{3}{2} + \sqrt{\frac{67}{2}}$ and $y = \frac{3}{2} - \sqrt{\frac{67}{2}}$
• $x$-intercepts: $x = 3 + \frac{\sqrt{161}}{2}$ and $x = 3 - \frac{\sqrt{161}}{2}$

Would you like more details or have any questions?

Related Questions

1. How do you find the center and radius of a circle from its equation?
2. Can we graph this circle to verify the intercepts?
3. How does the general formula of a circle help find intercepts?
4. What would be the intercepts if the radius changed?
5. How would changing the center affect the intercepts?

Tip

For any circle, remember that the equation is always in the form $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.