In the figure, the line 2t-by +9=0 cuts the circle at A and B on the .x-axis and y-axis respectively. If the centre of the circle lies on the line AB, find the equation of the circle.

To find the equation of the circle, given that the line \(2t - by + 9 = 0\) intersects the circle at points \(A\) and \(B\) on the \(x\)-axis and \(y\)-axis respectively, and the center of the circle lies on the line \(AB\), we can follow these steps:

### Step 1: Find the Coordinates of Points \(A\) and \(B\)

- **Point \(A\)** lies on the \(x\)-axis, so its \(y\)-coordinate is 0.
- **Point \(B\)** lies on the \(y\)-axis, so its \(x\)-coordinate is 0.

Given the line equation:

\[
2t - by + 9 = 0
\]

#### Finding \(A\) on the \(x\)-axis:
For \(A\), let \(y = 0\):

\[
2t + 9 = 0
\]
\[
t = -\frac{9}{2}
\]

So, the coordinates of \(A\) are \(\left(-\frac{9}{2}, 0\right)\).

#### Finding \(B\) on the \(y\)-axis:
For \(B\), let \(x = 0\):

\[
by = 2t + 9
\]
\[
y = \frac{2t + 9}{b}
\]

Using \(t = -\frac{9}{2}\):

\[
y = \frac{2\left(-\frac{9}{2}\right) + 9}{b} = \frac{-9 + 9}{b} = 0
\]

This suggests the point \(B\) is at \((0, 0)\), which we previously deduced may be incorrect unless the constant parameters \(t\) and \(b\) change. Therefore, we consider an alternate approach.

### Step 2: Equation of the Circle

The general form of the equation of a circle is:

\[
x^2 + y^2 + Dx + Ey + F = 0
\]

We know the circle passes through \(A = \left(-\frac{9}{2}, 0\right)\) and \(B = \left(0, \frac{9}{b}\right)\).

Since the center of the circle lies on the line \(AB\), the center \((h, k)\) must satisfy the equation of line \(AB\).

### Step 3: Find the Line \(AB\)

The slope of the line \(AB\) is:

\[
\text{slope of AB} = \frac{\frac{9}{b} - 0}{0 - \left(-\frac{9}{2}\right)} = \frac{9}{b} \times \frac{2}{9} = \frac{2}{b}
\]

The equation of the line \(AB\) in point-slope form is:

\[
y - 0 = \frac{2}{b}\left(x + \frac{9}{2}\right)
\]
\[
y = \frac{2}{b}x + \frac{9}{b}
\]

### Step 4: Use the Condition that the Center Lies on Line \(AB\)

Let the center of the circle be \((h, k)\). Since it lies on line \(AB\):

\[
k = \frac{2}{b}h + \frac{9}{b}
\]

The circle's equation becomes:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

Expanding it to match the general form:

\[
x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2
\]

Substituting \(k = \frac{2}{b}h + \frac{9}{b}\), you can further refine this based on \(A\) and \(B\) conditions to match:

\[
x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0
\]

### Conclusion

This equation represents a circle with its center on the line \(AB\). However, without specific numeric values for \(t\) and \(b\), the complete circle's equation is typically parametrized in terms of \(h\) and \(k\), relying on the further input that solidifies the specific location of the center and the radius.

Would you like more details or have any other questions? Here are 5 related questions:

1. How can we determine the center of a circle from its general equation?
2. What is the importance of the slope-intercept form in solving geometric problems?
3. How do the coordinates of intersecting points define a circle?
4. How can you find the radius of a circle given a chord and its midpoint?
5. What are the geometric properties of a line that bisects a circle?

**Tip:** When solving problems involving circles, always remember that the center's coordinates and radius can often be derived using intersection points and the geometric relationships they form with the circle's equation.

Learn how to find the equation of a circle when a line intersects it at points A and B on the x-axis and y-axis respectively, with the circle's center lying on the line AB. This problem involves applying the equation of a circle and geometric relationships. Suitable for high school students studying circle geometry.

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