Let C be the largest circle centred at (2, 0) and inscribed in the ellipse `x^2/36+y^2/16=1`.if `(1,alpha)` lies on C,then `10 albha^2` is equal to ……….

Answer

To solve the problem, we need to find the largest circle centered at (2, 0) that is inscribed in the given ellipse defined by the equation \( \frac{x^2}{36} + \frac{y^2}{16} = 1 \). We will then determine the value of \( 10 \alpha^2 \) where the point \( (1, \alpha) \) lies on this circle.

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse is given by:
\[ \frac{x^2}{36} + \frac{y^2}{16} = 1 \]
From this, we can identify:
- \( a^2 = 36 \) ⇒ \( a = 6 \)
- \( b^2 = 16 \) ⇒ \( b = 4 \)

### Step 2: Find the center and semi-axes of the ellipse
The center of the ellipse is at the origin (0, 0), and the semi-major axis (along the x-axis) is 6, while the semi-minor axis (along the y-axis) is 4.

### Step 3: Determine the largest inscribed circle
The largest circle that can be inscribed in the ellipse will be tangent to the ellipse at the points where the distance from the center of the ellipse to the ellipse is maximized. The circle is centered at (2, 0).

### Step 4: Calculate the distance from the center of the ellipse to the ellipse
The distance from the center of the ellipse (0, 0) to the center of the circle (2, 0) is:
\[ d = 2 \]
The radius of the ellipse along the x-axis at \( x = 2 \) can be found by substituting \( x = 2 \) into the ellipse equation:
\[ \frac{2^2}{36} + \frac{y^2}{16} = 1 \implies \frac{4}{36} + \frac{y^2}{16} = 1 \implies \frac{y^2}{16} = 1 - \frac{1}{9} = \frac{8}{9} \]
Thus,
\[ y^2 = \frac{128}{9} \implies y = \pm \frac{4\sqrt{2}}{3} \]
The maximum y-coordinate (the semi-minor axis) at \( x = 2 \) is \( \frac{4\sqrt{2}}{3} \).

### Step 5: Determine the radius of the inscribed circle
The radius \( r \) of the inscribed circle is the distance from the center of the circle (2, 0) to the ellipse along the y-axis:
\[ r = \frac{4\sqrt{2}}{3} \]

### Step 6: Find the coordinates of point \( (1, \alpha) \) on the circle
The equation of the circle centered at (2, 0) with radius \( r \) is:
\[ (x - 2)^2 + y^2 = r^2 \]
Substituting \( x = 1 \):
\[ (1 - 2)^2 + \alpha^2 = \left(\frac{4\sqrt{2}}{3}\right)^2 \]
This simplifies to:
\[ 1 + \alpha^2 = \frac{32}{9} \]
Thus,
\[ \alpha^2 = \frac{32}{9} - 1 = \frac{32}{9} - \frac{9}{9} = \frac{23}{9} \]

### Step 7: Calculate \( 10 \alpha^2 \)
Now we can calculate \( 10 \alpha^2 \):
\[ 10 \alpha^2 = 10 \times \frac{23}{9} = \frac{230}{9} \]

### Final Answer
Thus, \( 10 \alpha^2 \) is equal to \( \frac{230}{9} \).

---