Circle

Step 1: Concept
For the circle , radius .
Step 2: Analysis
; ; .
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Step 3: Conclusion
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Final Answer: (A)

Step 1: Concept
For the circle , radius .
Step 2: Analysis
; ; .
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Step 3: Conclusion
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Final Answer: (A)

Step 1: Concept
Equation of a circle with center and radius is .
Step 2: Analysis
Radius .
Equation: .
Step 3: Conclusion
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Final Answer: (B)

Step 1: Circle Through Origin
Equation: .
Step 2: Intercepts
x-intercept length .
y-intercept length .
Step 3: Center
The center is .
Step 4: Final Equations
Substitute and values:
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Final Answer: (a)

Step 1: General Circle
Let the center be . Circle passes through origin, so .
Step 2: Chord from Line
Distance from center to line : .
Step 3: Relationship
In (see diagram), .
Given chord , half-chord .
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Step 4: Locus
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Replacing center with , the center is . Thus .
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Final Answer: (b)

Step 1: Find the Center
The intersection point of the diameters is the center of the circle. Solving the equations and : Multiply the second equation by 2: . Subtract the first from this: . Substitute into : . The center is .
Step 2: Find the Radius
The circle passes through . Radius .
Step 3: Determine the Equation
The equation is . .
Final Answer: (c)

Step 1: Transformation Equations
For rotation of axes by angle :


Step 2: Substitution
Substitute into :
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Step 3: Expansion
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Step 4: Conclusion
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Since , the equation remains .
Final Answer: (d)

Step 1: Distance Between Lines
The two lines are parallel. The distance between them is the diameter of the circle.
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Step 2: Determine Radius
Diameter , so radius .
Step 3: Calculate Area
Area square unit.
Final Answer: (d)

Step 1: Identify Centers and Radii
Circle 1: , . Circle 2: , .
Step 2: Distance between Centers
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Step 3: Calculate the Angle
. Thus, .
Final Answer: (d)

Step 1: General Equation
Let the equation be (passing through origin, so ).
Step 2: Orthogonality Conditions
Condition: . With : . With : .
Step 3: Solve for Coefficients
Substitute : .
Step 4: Final Equation
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Final Answer: (b)

Step 1: Form the Equation
Substitute into the circle equation . .
Step 2: Simplify to Polynomial
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Step 3: Sum of Roots
For the quartic equation , the sum of roots is the negative of the coefficient of . Sum .
Final Answer: (a)

Step 1: Concept
The equation of the circle is .
Step 2: Analysis
Tangents at and are and .
Step 3: Evaluation
Squaring and adding both tangent equations:

.
Step 4: Conclusion
This simplifies to , which represents a circle.
Final Answer: (b)

Step 1: Equation of the circle.
The general equation of the circle is given by: where is the center of the circle, and the radius is given by:

Step 2: Find the condition for tangency.
Since the line is tangent to the circle at the point , the distance from the center of the circle to the line must equal the radius. The distance from a point to a line is given by: For the line , we rewrite it as .

Step 3: Apply the distance formula.
The distance from the center to the line is: Since the line is tangent, the distance from the center to the line is equal to the radius:

Step 4: Use the given condition.
We are also given that , so: Now, substituting into the equation, we solve for .

Step 5: Solve for .
After solving the equation, we find that the value of is , corresponding to option (B).

Step 6: Conclusion.
Thus, the value of is , corresponding to option (B).

Step 1: Understand the geometry of the problem.
We are given a circle that touches the x-axis and also touches another circle with center at and radius 2. We need to find the locus of the center of the circle.

Step 2: Analyze the situation geometrically.
The first circle touches the x-axis, meaning the distance from its center to the x-axis is equal to its radius, say . The second circle has a center at and radius 2.

Step 3: Use the distance condition.
The distance between the centers of the two circles is equal to the sum of their radii, i.e., the distance between the center of the first circle and is .

Step 4: Equation of the locus.
Let the center of the first circle be . The distance between and is given by the equation:

Step 5: Use the condition that the circle touches the x-axis.
Since the circle touches the x-axis, the distance from its center to the x-axis is equal to its radius . Therefore, the equation becomes:

Step 6: Conclusion.
By substituting into the distance equation, we obtain the equation of a parabola. Therefore, the locus of the center of the circle is a parabola, corresponding to option (A).

Step 7: Verification.
By analyzing the geometry and applying the distance conditions, we confirm that the locus is indeed a parabola.

Step 1: General equation of a circle.
The general equation of a circle is:
where the center is and the radius is .

Step 2: Use the fact that the center lies on the given line.
We are given that the center of the circle lies on the line . The center is , so substituting these into the line equation:

Step 3: Use the fact that the circle passes through the points (2, 3) and (4, 5).
Substitute the point into the general circle equation: Similarly, substitute the point :

Step 4: Solve the system of equations.
Now, we solve the system of equations (1), (2), and (3). Subtract Equation 2 from Equation 3: Now, substitute Equation 4 into Equation 1: Substitute into this equation: Substitute into Equation 4:

Step 5: Final substitution to find .
Now that we have and , substitute these values into Equation 2:


Step 6: Equation of the circle.
Thus, the equation of the circle is:

This matches option (C).

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}

Denominator =

Step 3: Final Answer

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}
for equilateral triangle





Since ,

Step 3: Final Answer

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}
By symmetry:




Step 3: Final Answer

Concept: Find centres and radii of circles, then use symmetry to determine rhombus diagonals.

Step 1: Find centres. First circle: Centre: , radius = 4 Second circle: Centre: , radius = 4

Step 2: Common region symmetric about y-axis. Distance between centres = 4

Step 3: Maximum rhombus formed: Diagonal along x-axis = 4 Other diagonal:

Step 4: Area of rhombus: Final Answer:

Concept: Circle touching both axes

Step 1: Centre of such circle = (r, r) = (2, 2).

Step 2: Distance between centres:

Step 3: Conclusion:
Distance = 5 units

Concept: Intercept form of circle gives center at midpoint of intercepts.

Step 1: Intercepts are 5 and 6. So center:

Step 2: Use origin to find radius: Final Answer:

Concept: Nearest point on a line to a circle lies along the perpendicular from the center of the circle.

Step 1:Convert circle to standard form
Divide by 16:

Step 2:Line equation

Step 3:Foot of perpendicular from center Here , Conclusion

Concept: Radical center = intersection of radical axes.

Step 1: Subtract equations.
Between first and second:

Step 2: Subtract first and third.


Step 3: Solve.
Solving gives:

Concept:
• Orthogonal circles:
• Tangent to line → radius = perpendicular distance

Step 1: Given circle.
Centre ,

Step 2: Let centre be .
Touches line :

Step 3: Apply orthogonality.


Step 4: Form locus.
Combining with symmetry in , locus reduces to:

Step 5: Final answer.

Step 1: Find center and radius.
Center , radius

Step 2: Slope of tangent.
Given line has slope Perpendicular slope So tangent:

Step 3: Use distance formula.
Distance from center to tangent = radius:

Step 4: Final equations.
Matching options:

Concept: Number of common tangents depends on the distance between centers relative to radii.

Step 1: Find centers and radii. Circle 1: → center , radius . Circle 2: → → center , radius .

Step 2: Distance between centers.

Step 3: Compare with . Circles touch externally. Number of common tangents = 3 (2 direct + 1 transverse).

Concept: Reflection about line :

Step 1:Replace , :

Step 2:Rearrange: Final:

Concept: Condition for perpendicular lines:

Step 1:Substitute in circle:

Step 2:Slopes of lines from origin: Using quadratic roots:

Step 3:Condition: Solving gives: