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Animated Solution for Mathematics - Circles: A variable circle passes through the fixed point $A (p, q)$ and touches x-axis. The locus of the other end of the diameter through $A$ is

Q: Why do we use the midpoint formula for the center?

Because $AB$ is the diameter! By definition, the center of a circle is the midpoint of any diameter. Since we defined $A$ and $B$ as the endpoints of the diameter, the center must be their midpoint.

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Visualized Solution

Visualizing the Setup

Fixed point $A (p, q)$
Variable circle passes through $A$
Circle touches the x-axis

Defining the Target Point $B (x, y)$

Let the other end of the diameter through $A$ be $B (x, y)$
We need to find the locus of $B$

Finding the Center $C$

Center $C$ is the midpoint of diameter $A B$
Midpoint formula: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$

Coordinates of Center $C$

$C = (\frac{x + p}{2}, \frac{y + q}{2})$

Calculating the Radius $r$

Radius $r = \frac{1}{2} A B$
Distance formula for $A B$ : $(x - p)^{2} + (y - q)^{2}$
$r = \frac{1}{2} (x - p)^{2} + (y - q)^{2}$

The Tangency Condition

The circle touches the x-axis
Distance from center $C$ to x-axis equals the radius $r$
$∣ y_{ce n t er} ∣ = r$

Equating the Radius

$y_{ce n t er} = \frac{y + q}{2}$
Equating both expressions for $r$ :
$∣ \frac{y + q}{2} ∣ = \frac{1}{2} (x - p)^{2} + (y - q)^{2}$

Squaring Both Sides

Square both sides to remove the square root and absolute value:
$(\frac{y + q}{2})^{2} = (\frac{1}{2} (x - p)^{2} + (y - q)^{2})^{2}$
$\frac{( y + q ) ^{2}}{4} = \frac{( x - p ) ^{2} + ( y - q ) ^{2}}{4}$

Canceling Denominators

Multiply both sides by $4$ :
$(y + q)^{2} = (x - p)^{2} + (y - q)^{2}$

Expanding the Squares

Expand $(y + q)^{2}$ and $(y - q)^{2}$ :
$y^{2} + q^{2} + 2 y q = (x - p)^{2} + y^{2} + q^{2} - 2 y q$

Simplifying the Equation

Cancel $y^{2}$ and $q^{2}$ from both sides:
$2 y q = (x - p)^{2} - 2 y q$

Isolating the Squared Term

Move $- 2 y q$ to the left side:
$2 y q + 2 y q = (x - p)^{2}$
$4 y q = (x - p)^{2}$

Final Locus Equation

Rearranging gives the final locus:
$(x - p)^{2} = 4 q y$
This represents a parabola opening upwards.

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

Let the fixed point be

A (p, q)

and the moving point at the other end of the diameter be

B (x, y)

The center

C

of the circle is the midpoint of the diameter

A B

. Therefore, the coordinates of the center are:

C = (\frac{x + p}{2}, \frac{y + q}{2})

The Geometric Constraint

Since the circle is tangent to the x-axis, the radius

r

of the circle must be equal to the absolute value of the y-coordinate of its center. Thus, we have:

r = \frac{y + q}{2}

We also know that the radius is half the length of the diameter

A B

. Using the distance formula for

A B

, we can express the radius as:

r = \frac{1}{2} (x - p)^{2} + (y - q)^{2}

The Master Equation

Equating the two expressions for the radius, we obtain:

\frac{1}{2} (x - p)^{2} + (y - q)^{2} = \frac{y + q}{2}

Squaring both sides of the equation yields:

\frac{( x - p ) ^{2} + ( y - q ) ^{2}}{4} = \frac{( y + q ) ^{2}}{4}

Multiplying both sides by

4

and expanding the squared terms, we get:

(x - p)^{2} + y^{2} - 2 y q + q^{2} = y^{2} + 2 y q + q^{2}

Final Result

By canceling the common terms

y^{2}

and

q^{2}

from both sides, the equation simplifies to:

(x - p)^{2} = 4 q y

This is the equation of a parabola. The locus of the point

B (x, y)

is a parabola with its vertex at

(p, 0)

and its axis parallel to the y-axis.

Common Traps & FAQs

Wait, why is the radius equal to the y-coordinate of the center? Shouldn't it be the x-coordinate?

Great question! When a circle touches the x-axis, the distance from the center to the x-axis is the vertical distance, which is the absolute value of the y-coordinate of the center. If it touched the y-axis, we would use the x-coordinate!

Why do we use the midpoint formula for the center?

Because

A B

is the diameter! By definition, the center of a circle is the midpoint of any diameter. Since we defined

A

and

B

as the endpoints of the diameter, the center must be their midpoint.

Is the locus always a parabola?

Yes! In this specific configuration, the geometric constraints force the point

B

to trace a parabolic path. It is beautiful how simple circle properties lead to conic sections, isn't it?

Animated Solution for Mathematics - Circles: A variable circle passes through the fixed point $A (p, q)$ and touches x-axis. The locus of the other end of the diameter through $A$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Target Point $B (x, y)$

Finding the Center $C$

Coordinates of Center $C$

Calculating the Radius $r$

The Tangency Condition

Equating the Radius

Squaring Both Sides

Canceling Denominators

Expanding the Squares

Simplifying the Equation

Isolating the Squared Term

Final Locus Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Constraint

The Master Equation

Final Result

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Target Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B(x,y)

Finding the Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Coordinates of Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Calculating the Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

The Tangency Condition

Equating the Radius

Squaring Both Sides

Canceling Denominators

Expanding the Squares

Simplifying the Equation

Isolating the Squared Term

Final Locus Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Constraint

The Master Equation

Final Result

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