Coordinate Geometry

Triangle Calculation

Let the points be defined as follows:

• A₁ =
• B₁ =
• C₁ =

Given and , the coordinates are:

• A₁ =
• B₁ =

Step 1: Calculate the length of OA₁

The distance is calculated as:

Simplify:

Step 2: Calculate the length of A₁B₁

The distance is given by:

Simplify:

Step 3: Verify the Orthocenter of △A₁B₁C₁

The orthocenter is the point where the altitudes of the triangle intersect. The altitudes are derived using the slopes of the sides:

• The slope of is undefined (vertical line), so its altitude passes through .
• The slopes of and confirm that the altitudes intersect at the origin .

Conclusion:

• The orthocenter is .

1. Understanding the Given Information:
We are given the equation and . From this, we need to find the value of . We are also given that , so we need to substitute this into the equation to find .

2. Substituting the Value of :
From the equation , substitute :

So, . Hence, the value of is .

3. Setting Up the Integral:
Next, we are given the equation for , the area:

4. Evaluating the Integral:
First, solve the integral :

Step 1: The integral of is . Evaluating it between and :

5. Completing the Area Calculation:
Now add the second part of the area formula: :

6. Final Calculation:
The total area is the sum of the two parts:

7. Concluding the Result:
We can express the final result as:

Final Answer:
The value of is .

To solve this problem, we need to find the correct matches between List-I and List-II based on the given conditions about the circle and the line.
We start with the equation of the line which is tangent to a circle centered at with radius . Given that , we need to calculate specific values for , , and the points and .
Step 1: Find the equation of the tangent. The line is tangent to the circle defined by the equation . For tangency, the perpendicular distance from the center to the line must be equal to the radius . The perpendicular distance is:

Setting this equal to , we get:

Step 2: Use the given condition .
Substitute the expression for :

Multiplying throughout by to eliminate the fraction:



After performing the necessary calculations, we find that: .
Step 3: Calculate using .
Substitute back into the equation :

Step 4: Determine and .
Since the line is tangent at , let the point be . For this point to lie on the circle and be tangent, the equation of the circle should hold true.
Simplifying, with and :
, solving this gives which implies , hence .
The point is diametrically opposite across the circle's center , using the midpoint formula:
is such that yielding: , hence .
Conclusion:

The correct option is thus , , , .