Coordinate Geometry

Step 1: Understand the Section Formula for External Division.
The section formula for external division is given by: where , , and are the points involved, and divides the line segment externally in the ratio .

Step 2: Apply the formula to the given points.
We are given: We will apply the section formula for the external division of the line segment to find the ratio in which divides externally.

Step 3: Use the formula to find the ratio.
The section formula for external division gives: Substituting the values of the coordinates of , , and , we calculate the ratio.

Step 4: Final Answer.
The ratio in which divides externally is .

Step 1: Identify centers and radii
The general equation of a circle is: Comparing with the given equations: For the first circle: Thus, the center is . For the second circle: Thus, the center is .

Step 2: Compute the angle between the circles
The angle between two circles is given by: Substituting values: Since , solving gives:

Step 3: Conclusion
Thus, the final answer is:

Step 1: Standard form of the given hyperbola
The given hyperbola equation is: Dividing by 232 to convert into standard form: Thus, the standard form is: where and .

Step 2: Condition for tangency
The equation of a general tangent to the hyperbola is: Comparing with the given line , we use the condition of tangency: Substituting values:

Step 3: Solving for
Solving the equation gives:

Step 4: Conclusion
Thus, the final answer is:

Step 1: Let the equation of the circle S be x² + y² + 2gx + 2fy + c = 0.
Since S cuts the given circles orthogonally, we have:

1. 2g(-1) + 2f(3) = c + 0 ⇒ -2g + 6f = c ...(1)
2. 2g(-2) + 2f(-1) = c + 6 ⇒ -4g - 2f = c + 6 ...(2)
3. 2g(-6) + 2f(1) = c + 3 ⇒ -12g + 2f = c + 3 ...(3)

Step 2: Solve the equations (1), (2), and (3).
From (1) and (2):
-2g + 6f = c
-4g - 2f = c + 6
Subtracting the equations:
2g + 8f = -6 ⇒ g + 4f = -3 ⇒ g = -3 - 4f ...(4)
From (1) and (3):
-2g + 6f = c
-12g + 2f = c + 3
Subtracting the equations:
10g + 4f = -3 ...(5)
Substitute (4) in (5):
10(-3 - 4f) + 4f = -3
-30 - 40f + 4f = -3
-36f = 27
f = - = -
Substitute f = - in (4):
g = -3 - 4(- ) = -3 + 3 = 0
Substitute g = 0 and f = - in (1):
c = -2(0) + 6(- ) = -
So, the equation of circle S is x² + y² - y - = 0.
2x² + 2y² - 3y - 9 = 0

Step 3: Find the equation of the tangent at (0, 3).
The equation of the tangent at (x₁, y₁) to x² + y² + 2gx + 2fy + c = 0 is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0.
Here, (x₁, y₁) = (0, 3), g = 0, f = - , c = - .
x(0) + y(3) + 0(x + 0) - (y + 3) - = 0
3y - y - - = 0
12y - 3y - 9 - 18 = 0
9y - 27 = 0
9y = 27
y = 3
Therefore, the equation of the tangent at (0, 3) on S = 0 is y = 3.

We are required to follow three transformations: 1. Translation of axes 2. Rotation of axes by 3. Reflection through the line Step 1: Translation of Axes
The point is translated when the origin is shifted to . Using the translation formula: Substituting the given point: Thus, the new point is .

Step 2: Rotation of Axes by
The rotation transformation formula is: Since , we have: So the new point is .

Step 3: Reflection through
The reflection transformation formula for reflection across is: Since and , the reflection gives:
Step 4: Final Answer

Final Answer: (C)

Step 1: Factorize the equation 6x² + 13xy + 6y² = 0.
6x² + 13xy + 6y² = 0
6x² + 9xy + 4xy + 6y² = 0
3x(2x + 3y) + 2y(2x + 3y) = 0
(3x + 2y)(2x + 3y) = 0
So, the two lines are 3x + 2y = 0 and 2x + 3y = 0.

Step 2: Find the intersection points of the lines.
Let the lines be:
L₁: 3x + 2y = 0
L₂: 2x + 3y = 0
L₃: x + 2y + 3 = 0
Intersection of L₁ and L₂:
3x + 2y = 0 and 2x + 3y = 0
Solving these, we get x = 0 and y = 0.
So, the intersection point is A(0, 0).
Intersection of L₁ and L₃:
3x + 2y = 0 and x + 2y + 3 = 0
Subtracting the equations:
2x - 3 = 0
x =
2y = -3x = -
y = -
So, the intersection point is B( , - ).
Intersection of L₂ and L₃:
2x + 3y = 0 and x + 2y + 3 = 0
From x + 2y + 3 = 0, x = -2y - 3.
Substitute in 2x + 3y = 0:
2(-2y - 3) + 3y = 0
-4y - 6 + 3y = 0
-y = 6
y = -6
x = -2(-6) - 3 = 12 - 3 = 9
So, the intersection point is C(9, -6).

Step 3: Calculate the area of the triangle.
The area of the triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by:
Area =
Area =
Area =
Area =
Area =
Therefore, the area of the triangle is square units.

Step 1: Find the intercepts of the line with the axes.
The equation of the line is 2x + 5y + a = 0.
Since the intercepts are with the positive coordinate axes, we must have a<0.
For x-intercept, put y = 0:
2x + a = 0
x = -
So, the x-intercept is A(- , 0).
For y-intercept, put x = 0:
5y + a = 0
y = -
So, the y-intercept is B(0, - ).

Step 2: Recognize the triangle formed.
The triangle formed by the line and the positive coordinate axes is a right-angled triangle with vertices A(- , 0), B(0, - ), and O(0, 0).

Step 3: Find the circumcenter and circumradius.
For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse AB.
Circumcenter =
Circumradius (R) is half the length of the hypotenuse AB.
AB =
Circumradius (R) =

Step 4: Use the given area of the circum-circle.
Area of the circum-circle =



|a| =
Therefore, |a| = 10.

We are given the curve and we need to find the area bounded by the curves , , and .

Step 1: Find the stationary point by differentiating the curve : Setting this equal to zero, we find , so .

Step 2: Calculate the area between the curves by setting up the definite integral between the points where the curves intersect.

Step 3: Using the standard integration techniques and solving the definite integrals, we obtain the area as . Thus, the correct answer is .

Step 1: Rotate the coordinate system. We are given the equation and we need to remove the -term by rotating the coordinate system. The angle of rotation is given by: where , , and . Substituting these values: Thus, .
Step 2: Apply the transformation. Using the formulas for coordinate rotation, we find the transformed equation:

Step 1: Understand the reflection property. When a ray of light reflects on the Y-axis, the x-coordinate of the incident ray changes sign, while the y-coordinate remains the same.
Let the incident point be A(2, 3) and the reflected point be B(3, 2).
Let the point of reflection on the Y-axis be P(a, b). Since P is on the Y-axis, a = 0.

Step 2: Use the reflection property to find the image of A.
The image of A(2, 3) with respect to the Y-axis is A'(-2, 3).

Step 3: Use the fact that A', P, and B are collinear.
Since A', P, and B are collinear, the slope of A'P is equal to the slope of PB.
Slope of A'P =
Slope of PB =
Since a = 0,
Slope of A'P =
Slope of PB =
Equating the slopes:

3(b - 3) = 2(2 - b)
3b - 9 = 4 - 2b
5b = 13

Step 4: Find the relationship between a and b. Since a = 0, we can write:
5b = 0 + 13
5b = a + 13
Therefore, 5b = a + 13.

Step 1: Let P be a point on x² + y² = 25. Let P be (x₁, y₁). Since P lies on x² + y² = 25, we have x₁² + y₁² = 25.

Step 2: Find the chord of contact L of P with respect to x² + y² = 9.
The equation of the chord of contact L is given by xx₁ + yy₁ = 9.

Step 3: Find the pole of the line L with respect to x² + y² = 36.
Let the pole be (h, k).
The equation of the polar of (h, k) with respect to x² + y² = 36 is hx + ky = 36.
This must be the same as the chord of contact L: xx₁ + yy₁ = 9.
Comparing coefficients, we have:

Thus, h = 4x₁ and k = 4y₁.
So, x₁ = and y₁ = .

Step 4: Find the locus of the pole (h, k).
Since x₁² + y₁² = 25, we substitute x₁ = and y₁ = :


h² + k² = 25 16 = 400
Thus, the locus of the pole (h, k) is x² + y² = 400.
Therefore, the locus of the poles of the lines L with respect to the circle x² + y² = 36 is x² + y² = 400.

Step 1: Find the center and radius of the circle.
The equation of the circle is x² + y² - 2x + 4y + 4 = 0.
Comparing with the general equation x² + y² + 2gx + 2fy + c = 0, we have:
2g = -2 \Rightarrow g = -1
2f = 4 \Rightarrow f = 2
c = 4
Center = (-g, -f) = (1, -2)
Radius (r) =
Let's check the distance: d =
Since d = and r = 1, the distance is greater than the radius, which is impossible. Let's check the given circle equation:
x² + y² - 2x + 4y + 4 = 0
(x-1)² - 1 + (y+2)² - 4 + 4 = 0
(x-1)² + (y+2)² = 1
Center (1, -2), radius r = 1.
Distance from center to chord:
d =
Again, d x = 1 - y
(1-y)² + y² - 2(1-y) + 4y + 4 = 0
1 - 2y + y² + y² - 2 + 2y + 4y + 4 = 0
2y² + 4y + 3 = 0
Let A(x₁, y₁) and B(x₂, y₂).
OA² = x₁² + y₁²
OB² = x₂² + y₂²
AB² = (x₂ - x₁)² + (y₂ - y₁)²
Let's use the cosine rule in triangle OAB:
AB² = OA² + OB² - 2(OA)(OB)

We are given the answer . Let

Therefore, the angle subtended by the chord at the origin is .

Step 1: Identify the center and radius of the circle
The given equation of the circle is: This represents a circle centered at with radius .

Step 2: Finding slopes of and
The points and lie on the circle. The slopes of the lines joining them to the origin (center of the circle) are:

Step 3: Find angle between and
Using the angle formula between two lines: where and ,

Step 4: Compute value

Step 5: Conclusion
Thus, the final answer is:

Step 1: Find the center and radius of the first circle.
The equation of the first circle is S = x² + y² - 14x + 6y + 33 = 0.
Comparing with the general equation x² + y² + 2gx + 2fy + c = 0, we have:
2g = -14 \Rightarrow g = -7
2f = 6 \Rightarrow f = 3
c = 33
Center C₁ = (-g, -f) = (7, -3)
Radius r₁ =

Step 2: Find the center and radius of the second circle.
The equation of the second circle is S' = x² + y² - a² = 0.
Comparing with the general equation x² + y² + 2gx + 2fy + c = 0, we have:
2g = 0 \Rightarrow g = 0
2f = 0 \Rightarrow f = 0
c = -a²
Center C₂ = (-g, -f) = (0, 0)
Radius r₂ = (since a )

Step 3: Find the distance between the centers.
Distance C₁ C₂ =

Step 4: Determine the condition for 4 common tangents.
For two circles to have 4 common tangents, they must be completely outside each other.
This means that the distance between the centers must be greater than the sum of the radii:
C₁ C₂ > r₁ + r₂


Since , we have:


Also, the circles should not intersect, so:
C₁ C₂ > |r₁ - r₂|


and
and
and
and
Since a , we have .
However, we need , so .
Since a , the only possible values are a = 1 and a = 2.
Thus, there are 2 possible values of a.
Therefore, the possible number of values of a is 2.

Step 1: Understanding given equation
The given equation states that the sum of reciprocals of distances and is given by: This equation can be transformed using the section formula in coordinate geometry.

Step 2: Finding intersection points
The given lines are: The parametric equation of a line passing through with slope is: Solving for and using this line equation, we find the required distances and .

Step 3: Solving for slope
After substituting in the distance equation and solving, we obtain:

Step 4: Conclusion
Thus, the final answer is:

Step 1: Find the foot of the perpendicular (a, b).
Let the point be P(3, 1) and the line be L: x + 3y + 4 = 0.
The slope of the line L is m_1 = - .
The slope of the line perpendicular to L is m_2 = - = 3.
The equation of the line passing through P(3, 1) and perpendicular to L is:
y - 1 = 3(x - 3)
y - 1 = 3x - 9
3x - y - 8 = 0
To find (a, b), solve the equations x + 3y + 4 = 0 and 3x - y - 8 = 0.
From x + 3y + 4 = 0, we get x = -3y - 4.
Substitute in 3x - y - 8 = 0:
3(-3y - 4) - y - 8 = 0
-9y - 12 - y - 8 = 0
-10y - 20 = 0
y = -2
x = -3(-2) - 4 = 6 - 4 = 2
So, (a, b) = (2, -2).

Step 2: Find the image (p, q) of (a, b) with respect to the line 3x - 4y + 11 = 0.
Let the line be M: 3x - 4y + 11 = 0.
The midpoint of (a, b) and (p, q) lies on the line M.
Midpoint =
Substitute in M:
3 - 4 + 11 = 0
3(p+2) - 4(q-2) + 22 = 0
3p + 6 - 4q + 8 + 22 = 0
3p - 4q + 36 = 0 ...(1)
The line joining (a, b) and (p, q) is perpendicular to M.
Slope of M =
Slope of the line joining (a, b) and (p, q) =
3(q+2) = -4(p-2)
3q + 6 = -4p + 8
4p + 3q - 2 = 0 ...(2)
Solve (1) and (2):
From (2), 3q = 2 - 4p, so q = .
Substitute in (1):
3p - 4 + 36 = 0
9p - 4(2-4p) + 108 = 0
9p - 8 + 16p + 108 = 0
25p + 100 = 0
p = -4
q =
So, (p, q) = (-4, 6).

Step 3: Calculate .

Therefore, .

Step 1: Compute the direction vector of the line The direction ratios of the line passing through points and are:

Step 2: Compute the normal vector of the plane The normal vector of the plane is found using the cross product of vectors formed by the three given points: Solving the determinant gives:

Step 3: Compute The angle between a line and a plane satisfies: Computing the dot product: Finding magnitudes:

Step 4: Compute