The locus of the mid-point of a chord of the circle x² + y² = 4 which subtends a right angle at the

Q: Wait, why is $\angle AOM = 45^\circ$? I thought the angle at the origin was $90^\circ$.

Great observation! The entire angle $\angle AOB$ is indeed $90^\circ$. However, because $\triangle OAB$ is an isosceles triangle (since $OA = OB = \text{radius}$), the line $OM$ drawn from the center to the midpoint of the base $AB$ acts as an angle bisector. Therefore, it splits the $90^\circ$ angle into two equal parts of $45^\circ$ each.

Q: Can we use the formula for the chord of contact to solve this?

You certainly could, but it might be overkill! The chord of contact formula is usually used when the chord is drawn from an external point. Here, we are looking for the locus of the midpoint of a chord that subtends a specific angle. The geometric approach using the properties of the triangle is much more elegant and less prone to algebraic errors.

Q: Why do we replace $(h,k)$ with $(x,y)$ at the end of the problem?

That is the standard procedure for finding a locus. We start by assuming the point is $(h,k)$ to distinguish it from the general coordinates of the plane. Once we find the relationship between $h$ and $k$ (the condition that must be satisfied for any such point), we replace them with $x$ and $y$ to express the final path or curve in the standard Cartesian form.

Question

The locus of the mid-point of a chord of the circle $x^2 + y^2 = 4$ which subtends a right angle at the origin is

Options:

(A) $x + y = 2$

(B) $x^2 + y^2 = 1$

(C) $x^2 + y^2 = 2$

(D) $x + y = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a circle defined by the equation $x^2 + y^2 = 4$. This circle is centered at the origin $(0,0)$ with a radius $r = 2$.

A chord $AB$ moves within this circle such that it subtends a right angle at the origin, meaning $\angle AOB = 90^{\circ}$. We aim to determine the locus of the midpoint $M(h,k)$ of this chord.

### The Geometric Insight

Consider the triangle $	riangle OAB$. Since $OA$ and $OB$ are radii of the circle, their lengths are equal to $2$, making $	riangle OAB$ an isosceles triangle.

Because $\angle AOB = 90^{\circ}$, $	riangle OAB$ is specifically an isosceles right-angled triangle. A fundamental theorem of geometry states that the line segment from the center of a circle to the midpoint of a chord is perpendicular to that chord.

Therefore, $OM \perp AB$. Since $OM$ is the altitude to the base of an isosceles triangle, it also bisects the vertex angle $\angle AOB$. Consequently, $\angle AOM = 45^{\circ}$.

### The Trigonometric Bridge

Focus on the right-angled triangle $	riangle OAM$. The hypotenuse $OA$ is the radius of the circle, which is $2$.

Using the definition of the cosine function:

$$\cos(45^{\circ}) = \frac{OM}{OA}$$

Substituting the known values:

$$\frac{1}{\sqrt{2}} = \frac{OM}{2}$$

Solving for $OM$, we find:

$$OM = \frac{2}{\sqrt{2}} = \sqrt{2}$$

This reveals that the distance from the origin to the midpoint $M$ is a constant value of $\sqrt{2}$, regardless of the chord's orientation.

### The Algebraic Conclusion

We translate this geometric property into an algebraic equation. The distance from the origin $(0,0)$ to the point $M(h,k)$ is given by:

$$OM = \sqrt{h^2 + k^2}$$

Equating this to our derived constant distance:

$$\sqrt{h^2 + k^2} = \sqrt{2}$$

Squaring both sides yields the equation of the locus:

$$h^2 + k^2 = 2$$

Replacing $(h,k)$ with the general coordinates $(x,y)$, the final locus is:

**$$x^2 + y^2 = 2$$**

This result represents a circle concentric with the original, possessing a radius of $\sqrt{2}$. The path of the midpoint is a perfect, elegant circle.

Animated Solution for Mathematics - Circles: The locus of the mid-point of a chord of the circle $x^{2} + y^{2} = 4$ which subtends a right angle at the origin is

Select Answer:

Visualized Solution

Equation of the Given Circle

Defining the Chord and Mid-point

The Right Angle Condition

Geometric Property of the Mid-point

Analyzing $△ O A B$

Trigonometry in $△ O A M$

Substituting Known Values

Calculating the Length of $O M$

Distance Formula for $O M$

Equating the Expressions

Squaring Both Sides

The Final Locus Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Geometric Insight

The Trigonometric Bridge

The Algebraic Conclusion

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the mid-point of a chord of the circle x2+y2=4 which subtends a right angle at the origin is