Circles

Centre of the circle:

Equation of diameter:

Now, equation of ellipse becomes:

Length of latus rectum (LR):

Thus,

We start with the given expression:

Now simplify step-by-step:

We know that . Hence, substituting and simplifying:

Now,

Substitute back into the main equation:

Therefore,

And,

Rewrite the Equation of the Circle:
The given equation of the circle is:
To find the center and radius, complete the square for and :
Completing the square:


So, the circle has center and radius .

Properties of the Inscribed Square:
Since the square is inscribed in the circle, its diagonal is equal to the diameter of the circle.
The diameter of the circle is , so the diagonal of the square is .

Calculate the Side Length of the Square:
For a square, if the diagonal length is , then the side length is given by:

Determine the Vertices of the Square:
Since one side of the square is parallel to the line , the square is oriented at a 45-degree angle. The center of the square is the same as the center of the circle, .

Using the center and the side length , the vertices of the square can be determined as:

Calculate :
Each vertex of the square satisfies:

Simplifying for each vertex and summing,
we get:

Coordinates of the centre will be:

Equation of circle:

Shortest distance:

Given:

To find the intersection point :

Since , we have:

The line passes through , so:

For circle : - Center lies on the y-axis with coordinates . - Given radius .

Applying the condition for tangency:

Squaring and simplifying:

Similarly, for circle : - Center lies on the y-axis at .

Calculating the square of the area of triangle :

Square of the area =

Step 1: Find the center of the circle.

The diameters intersect at the center of the circle. So, we solve the two linear equations:

Multiply the first equation by 3 and the second by 2 to eliminate : Subtracting:

Substitute in :

Therefore, the center of the circle is:

Step 2: Equation of the line joining given points

Let the two points be:

Slope of line :

Equation of line using point :

Simplify:

Hence, the equation of the line is:

Step 3: Condition for line to touch the circle (one point)

The line is tangent to the circle. The perpendicular distance from the center to this line equals the radius .

Distance from to :

Substitute :

Step 4: Equation of the circle

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Step 5: Find the point of tangency (P)

Equation of circle: Equation of tangent line:

Substitute in circle equation:

Simplify: \] \] Thus one point of contact .

Solve for :

Now :

Step 6: Find

Final Answer:

Let the center of the first circle be , with radius . The equation of the circle is: Now, the distance from the center of the circle to the line is the radius . The distance formula for a point to a line is: Substituting the values, we find the relationship between and . Similarly, for the second circle, we use the equation of the second line . The result of the calculations is the absolute difference between the squares of the radii:

First, rewrite the equation of the circle in standard form by completing the square for both and .

The equation is: Divide through by 4:

Now complete the square for and :

Simplifying:

Thus, the equation of the circle is:

The radius is the square root of , which is . Thus, the radius is .

Concept: If a line cuts a chord in a circle, the chord length can be computed using: where is the perpendicular distance from the center to the line.

Step 1: {Determine the centre and radius of the circle.} Since the circle touches the -axis at , the centre lies vertically above this point. Let centre be: Radius .

Step 2: {Use the intercept on the -axis.} Distance from centre to the -axis: Length of intercept on the -axis: Given: Thus centre:

Step 3: {Find perpendicular distance from centre to the line .} Line form: Distance:

Step 4: {Find the chord length.}