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Locus of Mid Point of the Chord of the Circle - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • 31 Questions around this concept.

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Circle $x^2+y^2=1$ intersect the x-axis at point B and y-axis at point A then finds the midpoint of AB.

Circle $x^2+y^2-2 x-2 y+1=0$ has a chord $A B=\sqrt{2}$ then find the Locus of the midpoint of chord AB

If a chord of the circle $x^2+y^2-4 x-2 y-c=0$ is trisected at the points $(1 / 3,1 / 3)$ and $(8 / 3,8 / 3)$, then

Two points P and Q are taken on the line joining the points A(0,0) and B(3 a, 0) such that A P=P Q=Q B. Circles are drawn on A P, P Q and Q B as diameters. The locus of the points, the sum of the squares of the tangents from which to the three circles is equal to $b^2$, is

The equations of two sides of a variable triangle are $\mathrm{x}=0$ and $\mathrm{y}=3$, and its third side is a tangent to parabola $y^2=6 x$. The locus of its circumcentre is:

A line segment $AB$ of length $\lambda$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $\lambda$. Then the locus of the point, that divides the line segment $AB$ in the ratio $2: 3$, is a circle of radius :

Let A be the point $(1,2)$ and B be any point on the curve $\mathrm{x}^2+y^2=16$. If the centre of the locus of the point P , which divides the line segment AB in the ratio $3: 2$ is the point $\mathrm{C}(\alpha, \beta)$ then the length of the line segment $A C$ is

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Find the sum of coefficients of x and y in the locus of the mid-point of the chord of the circle $x^2+y^2=a^2$ which subtends a right angle at the point $(p, q)$.

The locus of the mid points of the chords of the circle $x^2+y^2+4 x-6 y-12=0$ which subtend an angle of $\frac{\pi}{3}$ radians at its circumference is :

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 The locus of the mid-points of the chords of the circle \mathrm{x^{2}+y^{2}-2 x-6 y-10=0} which pass through the origin is:

Concepts Covered - 1

Locus of Mid Point of the Chord of the Circle

Locus of Mid Point of the Chord of the Circle

A circle with radius r,  centered at the point (h, k) and AB is its chord. Let M (x1 , y1) be the midpoint of the chord AB. 

From the figure,

$
\begin{aligned}
& \cos \theta=\frac{\mathrm{CM}}{r}=\frac{\sqrt{\left(\mathrm{x}_1-\mathrm{h}\right)^2+\left(\mathrm{y}_1-\mathrm{k}\right)^2}}{r} \\
& \Rightarrow 1-\sin ^2 \theta=\frac{\left(\mathrm{x}_1-\mathrm{h}\right)^2+\left(\mathrm{y}_1-\mathrm{k}\right)^2}{r^2} \\
& \Rightarrow \quad-\sin ^2 \theta=\frac{\left(\mathrm{x}_1-\mathrm{h}\right)^2+\left(\mathrm{y}_1-\mathrm{k}\right)^2-\mathrm{r}^2}{r^2}
\end{aligned}
$

$\therefore$ Required equation of locus is

$
\Rightarrow \quad \frac{(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2-\mathrm{r}^2}{r^2}=-\sin ^2 \theta
$
 

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Locus of Mid Point of the Chord of the Circle

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