Find the lengths of the intercepts of the circle x² + y²-5x-13y-14=0 on

the axes of co-ordinates.

To find the lengths of the intercepts of the circle on the coordinate axes, we first need to rewrite the equation of the circle in its standard form.

### Step 1: Rewriting the Equation in Standard Form

The given equation of the circle is:

\[
x^2 + y^2 - 5x - 13y - 14 = 0
\]

We'll complete the square for both \(x\) and \(y\) terms.

1. **For \(x\):**
   \[
   x^2 - 5x = \left(x^2 - 5x + \left(\frac{5}{2}\right)^2\right) - \left(\frac{5}{2}\right)^2 = \left(x - \frac{5}{2}\right)^2 - \frac{25}{4}
   \]

2. **For \(y\):**
   \[
   y^2 - 13y = \left(y^2 - 13y + \left(\frac{13}{2}\right)^2\right) - \left(\frac{13}{2}\right)^2 = \left(y - \frac{13}{2}\right)^2 - \frac{169}{4}
   \]

Now, substitute these into the original equation:

\[
\left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \left(y - \frac{13}{2}\right)^2 - \frac{169}{4} - 14 = 0
\]

Combine the constants on the right-hand side:

\[
\left(x - \frac{5}{2}\right)^2 + \left(y - \frac{13}{2}\right)^2 = \frac{25}{4} + \frac{169}{4} + 14
\]

Simplifying the constants:

\[
\frac{25}{4} + \frac{169}{4} + 14 = \frac{194}{4} + 14 = \frac{194}{4} + \frac{56}{4} = \frac{250}{4} = \frac{125}{2}
\]

Thus, the equation of the circle in standard form is:

\[
\left(x - \frac{5}{2}\right)^2 + \left(y - \frac{13}{2}\right)^2 = \frac{125}{4}
\]

### Step 2: Finding the Intercepts

- **X-intercepts**: Set \(y = 0\) and solve for \(x\).
  \[
  \left(x - \frac{5}{2}\right)^2 + \left(- \frac{13}{2}\right)^2 = \frac{125}{4}
  \]
  Simplify and solve for \(x\):
  \[
  \left(x - \frac{5}{2}\right)^2 + \frac{169}{4} = \frac{125}{4}
  \]
  \[
  \left(x - \frac{5}{2}\right)^2 = \frac{125}{4} - \frac{169}{4} = -\frac{44}{4} = -11
  \]
  Since a square can't be negative, there are **no x-intercepts**.

- **Y-intercepts**: Set \(x = 0\) and solve for \(y\).
  \[
  \left(- \frac{5}{2}\right)^2 + \left(y - \frac{13}{2}\right)^2 = \frac{125}{4}
  \]
  Simplify and solve for \(y\):
  \[
  \frac{25}{4} + \left(y - \frac{13}{2}\right)^2 = \frac{125}{4}
  \]
  \[
  \left(y - \frac{13}{2}\right)^2 = \frac{125}{4} - \frac{25}{4} = \frac{100}{4} = 25
  \]
  \[
  y - \frac{13}{2} = \pm 5
  \]
  So,
  \[
  y = \frac{13}{2} + 5 = \frac{13}{2} + \frac{10}{2} = \frac{23}{2} = 11.5
  \]
  or
  \[
  y = \frac{13}{2} - 5 = \frac{13}{2} - \frac{10}{2} = \frac{3}{2} = 1.5
  \]

So, the **y-intercepts** are at \(y = 11.5\) and \(y = 1.5\), and the lengths of the intercepts are:

- **X-intercepts**: None (as there are no x-intercepts).
- **Y-intercepts**: \(1.5\) and \(11.5\), so the length of the y-intercept is \(11.5 - 1.5 = 10\).

**Summary**:
- X-intercept: None
- Y-intercept length: \(10\) units

Would you like further details or have any questions?

### Related Questions:
1. How do you determine if a quadratic equation represents a circle?
2. What are the conditions for a circle to have intercepts on both axes?
3. How do you calculate the radius of a circle given its equation?
4. Can a circle have more than one intercept on a single axis?
5. What are the implications if the x-intercepts are imaginary?

**Tip**: When dealing with the equation of a circle, completing the square is often a crucial step to simplify and understand its geometry.

Learn how to find the lengths of intercepts of the circle x^2 + y^2 - 5x - 13y - 14 = 0 on the coordinate axes. This problem involves completing the square for circle equations and determining intercepts on the x and y axes. Suitable for high school students in Grades 10-12.

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