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What is the locus of the midpoints of all chords that can be drawn from a given point on a circle with a radius of 6.

Answer 1

The locus of the midpoints of all chords that can be drawn from a given fixed point $(a,b)$ on a circle with a radius of 6 units, is a circle of radius 3 units with center at a point whose x & y coordinates are shifted from the center of the given circle by $\frac{a}{2}$ and $\frac{b}{2}$ respectively.

Given: A circle of radius 6 units

To find: The locus of the midpoint of all chords that can be drawn from a given point on the circle.

To find the required locus, we need to know the following:

• Locus of a moving point is the trajectory of that point. It is the geometrical figure represented by the equation which is satisfied by the coordinates of the moving point.

• A chord of a circle is a line segment joining any points of a circle.

• Equation of a circle with center at origin and radius of $r$ units is $x^{2} +y^{2} =r^{2}$
• According to the midpoint formula, the coordinates of the midpoint of the line segment joining the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $(\frac{x_{1}+x_{2} }{2} ,\frac{y_{1}+y_{2} }{2} )$

Let, without loss of generality, the given circle be centered at the origin. Even if it is not, we can shift the origin to the center of the given circle with coordinate transformation.

Then, the equation of the given circle is $x^{2}+y^{2} =6^{2}$, that is, $x^{2}+y^{2} = 36$

Let the coordinates of the given fixed point be $(a,b)$

Let the coordinates of any point on the circle be $(p,q)$ and let the coordinates of the midpoint of the chord joining the points $(a,b)$ and $(p,q)$ be $(h,k)$

We have to find the locus of $(h,k)$

Then, using the midpoint formula,

$(h,k)=(\frac{a+p}{2} ,\frac{b+q}{2})$

On solving, we get,

$p=2h-a,q=2k-b$

Since $(a,b)$ and $(p,q)$ are both points on the given circle, they satisfy the equation of the circle, $x^{2}+y^{2} = 36$

Then,

$a^{2} +b^{2} =36$

$p^{2} +q^{2} =36$

Substituting $p=2h-a,q=2k-b$ in $p^{2} +q^{2} =36$, we have,

$(2h-a)^{2} +(2k-b)^{2} =36$

$(2(h-\frac{a}{2}) )^{2} +(2(k-\frac{b}{2}))^{2} =36$

$4(h-\frac{a}{2})^{2} +4(k-\frac{b}{2})^{2} =36$

$(h-\frac{a}{2})^{2} +(k-\frac{b}{2})^{2} =\frac{36}{4}$

$(h-\frac{a}{2})^{2} +(k-\frac{b}{2})^{2} =9$

$(h-\frac{a}{2})^{2} +(k-\frac{b}{2})^{2} =3^{2}$

This is the locus of the point $(h,k)$

Replace $(h,k)=(x,y)$ to get,

$(x-\frac{a}{2})^{2} +(y-\frac{b}{2})^{2} =3^{2}$

This is the equation of a circle with center at $(\frac{a}{2} ,\frac{b}{2} )$ and radius 3 units.

Thus, we can conclude that the locus of the midpoints of all chords that can be drawn from a given fixed point $(a,b)$ on a circle with a radius of 6 units, is a circle of radius 3 units with center at a point whose x & y coordinates are shifted from the center of the given circle by $\frac{a}{2}$ and $\frac{b}{2}$ respectively.

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